Class 12 Maths Differential Equations – NCERT Formulas & Examples

 

Class 12 Maths Chapter 9 – Differential Equations

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Introduction

Differential equations form the backbone of higher mathematics. They describe relationships between functions and their rates of change — essential for understanding motion, growth, and natural processes. In Class 12, you’ll learn how to form, solve, and apply differential equations, moving from basic NCERT concepts to IIT-level analytical thinking.

Formulas and Key Concepts

• Derivative: \( \frac{dy}{dx} \)
• Order: The order of a differential equation is the order of the highest derivative present.
• Degree: The degree is the power of the highest derivative (after making it free from radicals/fractions).
• General Solution: The family of all possible solutions containing arbitrary constants.
• Particular Solution: A solution obtained by assigning specific values to constants.
• Types of Differential Equations:
  • Variables separable
  • Homogeneous
  • Linear (first order)
  • Exact equations

Solved Examples

Example 1: Solve \( \frac{dy}{dx} = 3x^2 \)

Integrate both sides with respect to \(x\): \[ y = \int 3x^2 dx = x^3 + C \] Hence, the solution is \( y = x^3 + C \).

Example 2: Solve \( \frac{dy}{dx} + y = e^x \)

This is a linear differential equation. Integrating factor (I.F) = \( e^{\int 1 dx} = e^x \). Multiply throughout by I.F: \[ e^x \frac{dy}{dx} + e^x y = e^{2x} \] Integrate both sides: \[ \frac{d}{dx}(y e^x) = e^{2x} \Rightarrow y e^x = \frac{e^{2x}}{2} + C \] So, \( y = \frac{e^x}{2} + Ce^{-x} \).

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NCERT + IIT Level Questions with Solutions (80 in total)

Below are progressive questions from NCERT to IIT-JEE advanced level. Click “Show Solution” to view each detailed explanation.

Q1. Find the general solution of \( \frac{dy}{dx} = y \tan x \)

Separate the variables: \[ \frac{dy}{y} = \tan x \, dx \] Integrate both sides: \[ \ln y = -\ln|\cos x| + C \] Hence, \( y = C \sec x \).

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50 FAQs on Differential Equations

These frequently asked questions range from simple conceptual doubts to IIT-level problem-solving approaches.

FAQ 1: What is the difference between a general and particular solution?

A general solution contains arbitrary constants representing a family of curves, while a particular solution is obtained by substituting given conditions to find specific values of constants.

FAQ 2: What makes an equation linear in \(y\)?

An equation is linear in \(y\) if it can be written in the form \[ \frac{dy}{dx} + Py = Q \] where \(P\) and \(Q\) are functions of \(x\).

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Related Articles

• Class 12 Relations and Functions
• Class 12 Integrals

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Q1. Find the general solution of \( \frac{dy}{dx} = y \tan x \)

Separate variables:
\(\frac{dy}{y} = \tan x \, dx\)
Integrate both sides:
\(\ln y = -\ln|\cos x| + C\)
Hence, \(y = C \sec x\).

Q2. Solve \( \frac{dy}{dx} = e^{x-y} \)

Rewrite as \( e^y \, dy = e^x \, dx \)
Integrate both sides:
\(\int e^y dy = \int e^x dx\)
\(\Rightarrow e^y = e^x + C\)
\(\Rightarrow y = \ln(e^x + C)\).

Q3. Solve \( \frac{dy}{dx} = \frac{x + y}{x} \)

Rewrite: \(\frac{dy}{dx} - \frac{y}{x} = 1\)
This is a linear DE. Integrating factor (I.F) = \( e^{\int -1/x \, dx} = \frac{1}{x} \)
Multiply by I.F:
\(\frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = \frac{1}{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{1}{x}\)
Integrate: \(\frac{y}{x} = \ln|x| + C\)
Hence, \( y = x(\ln|x| + C) \).

Q4. Solve \( \frac{dy}{dx} = 1 + y^2 \)

Separate variables:
\(\frac{dy}{1+y^2} = dx\)
Integrate: \(\tan^{-1}y = x + C\)
Hence, \( y = \tan(x + C) \).

Q5. Solve \( \frac{dy}{dx} + y = \sin x \)

This is a linear DE.
I.F = \( e^{\int 1 \, dx} = e^x \)
Multiply by I.F:
\( e^x \frac{dy}{dx} + e^x y = e^x \sin x \)
\(\Rightarrow \frac{d}{dx}(ye^x) = e^x \sin x \)
Integrate by parts:
\(\int e^x \sin x \, dx = \frac{e^x}{2}(\sin x - \cos x)\)
Hence, \( y e^x = \frac{e^x}{2}(\sin x - \cos x) + C \)
\(\Rightarrow y = \frac{1}{2}(\sin x - \cos x) + Ce^{-x} \).

Q6. Find the particular solution of \( \frac{dy}{dx} = xy^2 \), given that \( y = 1 \) when \( x = 0 \).

Separate variables:
\(\frac{dy}{y^2} = x \, dx\)
Integrate: \(-\frac{1}{y} = \frac{x^2}{2} + C\)
Using \(x=0, y=1\): \(-1 = 0 + C \Rightarrow C = -1\)
Hence, \(-\frac{1}{y} = \frac{x^2}{2} - 1\)
\(\Rightarrow y = \frac{1}{1 - \frac{x^2}{2}} = \frac{2}{2 - x^2}\).

Q7. Solve \( (x + y) \, dx = (x - y) \, dy \)

Rearrange: \(\frac{dy}{dx} = \frac{x + y}{x - y}\)
Let \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute: \( v + x\frac{dv}{dx} = \frac{1 + v}{1 - v} \)
Simplify: \( x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v(1 - v)}{1 - v} = \frac{1 + v^2}{1 - v} \)
Separate: \(\frac{1 - v}{1 + v^2} \, dv = \frac{dx}{x}\)
Integrate both sides:
\(\int \frac{1 - v}{1 + v^2} \, dv = \ln|x| + C\)
\(\Rightarrow \tan^{-1}v - \frac{1}{2}\ln(1 + v^2) = \ln|x| + C\)
Substitute back \(v = y/x\): \(\tan^{-1}\frac{y}{x} - \frac{1}{2}\ln(x^2 + y^2) = \ln|x| + C\).

Q8. Find the general solution of \( \frac{dy}{dx} = e^{x-y} + y \)

Let \( v = e^y \Rightarrow \frac{dv}{dx} = e^y \frac{dy}{dx} = v(e^{x-y} + y) = v(e^x/v + y) = e^x + vy \)
This becomes \(\frac{dv}{dx} - vy = e^x\), which is not directly separable. We’ll solve numerically or by approximation for higher-level practice (covered in IIT section).

Q9. Solve \( \frac{dy}{dx} = \frac{x^2 + y^2}{xy} \)

Let \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute:
\( v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{xvx} = \frac{1 + v^2}{v} \)
Simplify: \( x\frac{dv}{dx} = \frac{1 + v^2}{v} - v = \frac{1}{v} \)
Separate: \( v \, dv = \frac{dx}{x} \)
Integrate: \( \frac{v^2}{2} = \ln|x| + C \Rightarrow v^2 = 2\ln|x| + K \)
Hence, \( y^2 = x^2(2\ln|x| + K) \).

Q10. Solve \( \frac{dy}{dx} + y\tan x = \sin x \)

This is a linear DE. I.F = \( e^{\int \tan x \, dx} = e^{-\ln|\cos x|} = \sec x \)
Multiply by I.F: \(\sec x \frac{dy}{dx} + y\sec x \tan x = \sec x \sin x = \tan x\)
\(\Rightarrow \frac{d}{dx}(y \sec x) = \tan x\)
Integrate: \( y \sec x = \int \tan x \, dx = -\ln|\cos x| + C \)
Hence, \( y = \cos x(-\ln|\cos x| + C) \).

Q11. Solve: \(\displaystyle \frac{dy}{dx} - \frac{2}{x}y = 1\).

This is a linear equation of the form \( \dfrac{dy}{dx}+P(x)y=Q(x)\) with \(P(x)=-\dfrac{2}{x}\).

I.F. \(= e^{\int -2/x\,dx} = x^{-2}.\)

Multiply through by I.F.: \(\dfrac{d}{dx}\left(y x^{-2}\right) = x^{-2}.\)

Integrate: \(y x^{-2} = \int x^{-2} dx = -x^{-1} + C.\)

So \(y = -x + C x^{2}.\)

Q12. Solve the Bernoulli equation: \(\displaystyle \frac{dy}{dx} + y = y^2.\)

Write as Bernoulli with \(n=2\). Divide both sides by \(y^2\):

\(y^{-2}\frac{dy}{dx} + y^{-1} = 1.\)

Set \(v = y^{-1}\). Then \(dv/dx = -y^{-2} dy/dx\). So the equation becomes

\(-dv/dx + v = 1 \Rightarrow dv/dx - v = -1.\)

I.F. \(= e^{\int -1\,dx} = e^{-x}.\)

\(\dfrac{d}{dx}(v e^{-x}) = -e^{-x}.\)

Integrate: \(v e^{-x} = \int -e^{-x} dx = e^{-x} + C.\)

So \(v = 1 + C e^{x}\) and \(y = \dfrac{1}{v} = \dfrac{1}{1 + C e^{x}}\).

Q13. Solve the exact equation: \((2xy + \cos x)\,dx + (x^2 + 2y)\,dy = 0.\)

Check exactness: \(M(x,y)=2xy+\cos x\) so \(M_y=2x.\) \(N(x,y)=x^2+2y\) so \(N_x=2x.\) They are equal, so exact.

Find potential \( \Phi(x,y)\) with \(\Phi_x = M\):

\(\Phi = \int (2xy+\cos x)\,dx = x^2 y + \sin x + h(y).\)

Differentiate w.r.t. \(y\): \(\Phi_y = x^2 + h'(y) = N = x^2 + 2y.\) So \(h'(y)=2y\Rightarrow h(y)=y^2.\)

Solution: \(x^2 y + \sin x + y^2 = C.\)

Q14. Solve by separation: \(\displaystyle \frac{dy}{dx} = \frac{x^2 + 1}{y^2 + 1}.\)

Separate variables: \((y^2+1)\,dy = (x^2+1)\,dx.\)

Integrate both sides:

\(\int (y^2+1)\,dy = \int (x^2+1)\,dx \Rightarrow \dfrac{y^3}{3} + y = \dfrac{x^3}{3} + x + C.\)

That is the implicit general solution: \(\dfrac{y^3}{3}+y - \dfrac{x^3}{3} - x = C.\)

Q15. Find the orthogonal trajectories of the family \(y = k x^2\) (where \(k\) is a parameter).

Differentiate the family: \(y = kx^2 \Rightarrow dy/dx = 2kx.\)

But \(k = y/x^2\), so \(dy/dx = 2\cdot \dfrac{y}{x}.\)

Slope of orthogonal trajectories is negative reciprocal: \(m_{\perp} = -\dfrac{1}{2y/x} = -\dfrac{x}{2y}.\)

Solve differential equation: \(\dfrac{dy}{dx} = -\dfrac{x}{2y}\).

Separate: \(2y\,dy = -x\,dx\). Integrate:

\(y^2 = -\dfrac{x^2}{2} + C\) or \( \boxed{y^2 + \dfrac{x^2}{2} = C} \) — family of orthogonal trajectories.

Q16. Model the population growth with \(\dfrac{dy}{dt} = ky\) and find \(y(t)\) given \(y(0)=y_0\).

Separate: \(\dfrac{dy}{y} = k\,dt.\)

Integrate: \(\ln|y| = kt + C.\)

Exponentiate: \(y = Ce^{kt}.\) Use \(y(0)=y_0\Rightarrow C=y_0.\)

So the solution is \( \boxed{y(t)=y_0 e^{kt}}.\)

Q17. Solve: \(\displaystyle \frac{dy}{dx} + 2x y = x.\)

This is linear with \(P(x)=2x\). I.F. \(= e^{\int 2x\,dx} = e^{x^2}.\)

Multiply through: \(\dfrac{d}{dx}(y e^{x^2}) = x e^{x^2}.\)

Integrate RHS: \(\int x e^{x^2} dx = \dfrac{1}{2} e^{x^2} + C.\)

So \(y e^{x^2} = \dfrac{1}{2} e^{x^2} + C \Rightarrow y = \dfrac{1}{2} + C e^{-x^2}.\)

Q18. Solve the Clairaut-type equation \(y = x p + p^2\) where \(p=\dfrac{dy}{dx}\). Find the general and singular solution.

Clairaut form \(y = xp + f(p)\) has general solution \(y = Cx + f(C)\) (straight lines) and a singular envelope found by eliminating \(C\).

Here \(f(p)=p^2\). General solution: \(y = Cx + C^2\).

For singular solution, differentiate the original equation w.r.t. \(x\): \(p = p + x dp/dx + 2p dp/dx \Rightarrow (x+2p)dp/dx = 0.\)

So either \(dp/dx=0\) (gives general solution) or \(x + 2p = 0 \Rightarrow p = -\dfrac{x}{2}.\)

Plug into original: \(y = x\left(-\dfrac{x}{2}\right) + \left(-\dfrac{x}{2}\right)^2 = -\dfrac{x^2}{2} + \dfrac{x^2}{4} = -\dfrac{x^2}{4}.\)

So singular solution (envelope) is \( \boxed{y = -\dfrac{x^2}{4}}.\)

Q19. Solve: \(\displaystyle \frac{dy}{dx} = \frac{x^2 + y^2}{2xy}.\)

This is homogeneous. Put \(y = vx \Rightarrow dy/dx = v + x dv/dx.\)

Substitute: \(v + x\frac{dv}{dx} = \dfrac{1 + v^2}{2v}.\)

Simplify: \(x\frac{dv}{dx} = \dfrac{1 + v^2}{2v} - v = \dfrac{1 - v^2}{2v}.\)

Separate: \(\dfrac{2v}{1 - v^2} dv = \dfrac{dx}{x}.\)

Integrate left: let \(u=1-v^2\), \(du=-2v dv\) so \(\int \dfrac{2v}{1-v^2} dv = -\int \dfrac{du}{u} = -\ln|1-v^2|.\)

Thus \(-\ln|1-v^2| = \ln|x| + C\Rightarrow \ln|1-v^2| = -\ln|x| + C'.\)

So \(1 - v^2 = \dfrac{K}{x}\) where \(K\) is constant. Replace \(v=\dfrac{y}{x}\):

\(1 - \dfrac{y^2}{x^2} = \dfrac{K}{x} \Rightarrow y^2 = x^2 - Kx.\)

That is the implicit solution: \(\boxed{y^2 = x^2 - Kx}.\)

Q20. Solve the exact equation: \((y - x)\,dx + (x + y)\,dy = 0.\)

Check exactness: \(M=y-x \Rightarrow M_y=1.\) \(N=x+y \Rightarrow N_x=1.\) Exact.

Find potential function \(\Phi\) with \(\Phi_x = y-x\):

\(\Phi = \int (y-x)\,dx = xy - \dfrac{x^2}{2} + h(y).\)

Differentiate wrt \(y\): \(\Phi_y = x + h'(y) = N = x + y \Rightarrow h'(y)=y \Rightarrow h=\dfrac{y^2}{2}.\)

Solution: \(xy - \dfrac{x^2}{2} + \dfrac{y^2}{2} = C.\)

Q21. Solve: \(\displaystyle \frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\).

Homogeneous. Put \(y = vx\), so \(dy/dx = v + x dv/dx\). Substitute:

\(v + x\frac{dv}{dx} = \dfrac{v^2 - 1}{2v}.\)

\(x\frac{dv}{dx} = \dfrac{v^2 - 1}{2v} - v = -\dfrac{1+v^2}{2v}.\)

Separate: \(\dfrac{2v}{1+v^2} dv = -\dfrac{dx}{x}.\)

Integrate: \(\ln(1+v^2) = -\ln|x| + C.\) So \(1+v^2 = \dfrac{K}{x}.\)

Replace \(v=\dfrac{y}{x}\): \(1 + \dfrac{y^2}{x^2} = \dfrac{K}{x} \Rightarrow y^2 = Kx - x^2.\)

Q22. Solve the linear equation: \(\displaystyle \frac{dy}{dx} + y \tan x = \sin x \cos x.\)

Linear with \(P(x)=\tan x\). I.F. \(= e^{\int \tan x dx} = \sec x.\)

Multiply: \(\dfrac{d}{dx}(y\sec x) = \sin x\cos x \sec x = \sin x.\)

Integrate: \(y\sec x = -\cos x + C.\)

So \(y = -\cos^2 x + C\cos x\). (You can write \(-\cos^2 x = \sin^2 x -1\) if preferred.)

Q23. Solve: \(\displaystyle \frac{dy}{dx} = \frac{x + 2y}{x - 2y}\).

Put \(y = vx \Rightarrow dy/dx = v + x dv/dx\).

Substitute: \(v + x\frac{dv}{dx} = \dfrac{1 + 2v}{1 - 2v}.\)

\(x\frac{dv}{dx} = \dfrac{1 + 2v}{1 - 2v} - v = \dfrac{1 + 2v - v(1 - 2v)}{1 - 2v} = \dfrac{1 + v^2}{1 - 2v}.\)

Separate: \(\dfrac{1 - 2v}{1 + v^2} dv = \dfrac{dx}{x}.\)

Integrate: \(\int \dfrac{1 - 2v}{1 + v^2} dv = \ln|x| + C.\)

Left splits: \(\tan^{-1}v - \ln\sqrt{1+v^2} - \ln|1+v^2|^{?}\) — to keep it compact, write the integrated form

\(\tan^{-1}v - \ln(1+v^2) - 2\tan^{-1}v\) — better to present final implicit solution:

\(\tan^{-1}\!\left(\frac{y}{x}\right) - \frac{1}{2}\ln\!\big(1 + (y/x)^2\big) - 2\ln|x| = C.\)

(This is an implicit relation; you can simplify constants differently.)

Q24. Solve the Bernoulli equation: \(\displaystyle \frac{dy}{dx} - \frac{y}{x} = x^2 y^3.\)

Bernoulli with \(n=3\). Divide by \(y^3\): \(y^{-3} dy/dx - \dfrac{y^{-2}}{x} = x^2.\)

Let \(v = y^{-2} \Rightarrow dv/dx = -2 y^{-3} dy/dx.\) So \( -\frac{1}{2} dv/dx - \dfrac{v}{x} = x^2.\)

Multiply by -2: \(dv/dx + \dfrac{2v}{x} = -2x^2.\)

I.F. \(= e^{\int 2/x dx} = x^2.\) Multiply: \(\dfrac{d}{dx}(v x^2) = -2x^4.\)

Integrate: \(v x^2 = -\dfrac{2x^5}{5} + C.\) Thus \(v = -\dfrac{2x^3}{5} + \dfrac{C}{x^2}.\)

Return: \(y^{-2} = -\dfrac{2x^3}{5} + \dfrac{C}{x^2}\Rightarrow y = \pm\frac{1}{\sqrt{\dfrac{C}{x^2} - \dfrac{2x^3}{5}}}.\)

Q25. Solve the exact equation: \((e^{x} \sin y + 2x)\,dx + (e^{x} \cos y + 3y^2)\,dy = 0.\)

Check exactness: \(M=e^x \sin y + 2x \Rightarrow M_y = e^x \cos y.\) \(N = e^x \cos y + 3y^2 \Rightarrow N_x = e^x \cos y.\) Exact.

Find potential \(\Phi\) via \(\Phi_x = M\):

\(\Phi = \int (e^x \sin y + 2x) dx = e^x \sin y + x^2 + h(y).\)

\(\Phi_y = e^x \cos y + h'(y) = N = e^x \cos y + 3y^2\Rightarrow h'(y)=3y^2.\)

\(h(y)=y^3.\) So solution: \(e^x \sin y + x^2 + y^3 = C.\)

Q26. Solve: \(\displaystyle \frac{dy}{dx} = \frac{2x + y}{x - 2y}\) with substitution to make it homogeneous.

Set \(y = vx\). Then \(dy/dx = v + x dv/dx\). Substitute:

\(v + x\frac{dv}{dx} = \dfrac{2 + v}{1 - 2v}.\)

\(x\frac{dv}{dx} = \dfrac{2 + v}{1 - 2v} - v = \dfrac{2 + v - v(1 - 2v)}{1 - 2v} = \dfrac{2 + 2v^2}{1 - 2v}.\)

Separate: \(\dfrac{1 - 2v}{1 + v^2} dv = \dfrac{2\,dx}{x}.\)

Integrate and simplify to get implicit relation: \(\tan^{-1}v - \ln(1+v^2) = 2\ln|x| + C.\) Replace \(v=y/x\) to finish.

Q27. Solve: \(\displaystyle \frac{dy}{dx} = \frac{1 + x^2}{2xy} - \frac{y}{x}.\)

Rearrange: \(\frac{dy}{dx} + \frac{y}{x} = \dfrac{1 + x^2}{2x y}.\) Multiply by \(2y\): \(2y\frac{dy}{dx} + \frac{2y^2}{x} = \dfrac{1+x^2}{x}.\)

Note left is derivative: \(\dfrac{d}{dx}(y^2) + \dfrac{2y^2}{x} = \dfrac{1+x^2}{x}.\)

Let \(u = y^2\). Then \(du/dx + \dfrac{2u}{x} = \dfrac{1+x^2}{x}.\)

I.F. \(= e^{\int 2/x dx} = x^2.\) Multiply: \(\dfrac{d}{dx}(u x^2) = x + x^3.\)

Integrate: \(u x^2 = \dfrac{x^2}{2} + \dfrac{x^4}{4} + C.\) So \(y^2 = \dfrac{1}{2} + \dfrac{x^2}{4} + \dfrac{C}{x^2}.\)

Q28. Solve the Riccati-type (reducible) equation: \(\displaystyle \frac{dy}{dx} = x^2 + y^2\) by substitution \(y = \frac{1}{u}\frac{du}{dx}\) if applicable, or comment on solvability.

This Riccati-type equation \(dy/dx = y^2 + x^2\) does not have a straightforward elementary general solution. A standard approach is to look for a particular solution or use substitution when a particular solution is known. Without a known particular solution, the general solution requires special functions or series.

For exams, state: not solvable in elementary closed form; use numerical methods or series. If a particular solution \(y_p\) is found, reduce to linear via \(y = y_p + 1/u\).

Q29. Solve: \(\displaystyle \frac{dy}{dx} = \frac{y}{x} + \frac{x^2}{y}\).

Let \(u = y^2\). Then \(du/dx = 2y dy/dx.\) Multiply the DE by \(2y\):

\(2y\frac{dy}{dx} = \frac{2y^2}{x} + 2x^2.\) So \(du/dx = \dfrac{2u}{x} + 2x^2.\)

This is linear in \(u\): \(du/dx - \dfrac{2u}{x} = 2x^2.\)

I.F. \(= e^{\int -2/x dx} = x^{-2}.\) Multiply: \(\dfrac{d}{dx}(u x^{-2}) = 2.\)

Integrate: \(u x^{-2} = 2x + C \Rightarrow u = 2x^3 + C x^2.\)

Thus \(y^2 = 2x^3 + C x^2 \Rightarrow y = \pm x\sqrt{2x + C}.\)

Q30. Find the general solution of the equation with integrating factor dependent on x: \(\displaystyle (x\ln x)\,dy + (y + x)\,dx = 0.\)

Write in standard form: \((y+x)dx + (x\ln x)dy = 0.\) Check exactness: \(M=y+x \Rightarrow M_y=1.\) \(N=x\ln x \Rightarrow N_x = \ln x + 1.\) Not exact.

Test for µ(x): \(\dfrac{M_y - N_x}{N} = \dfrac{1 - (\ln x +1)}{x\ln x} = -\dfrac{\ln x}{x\ln x} = -\dfrac{1}{x}.\) This depends only on x, so integrating factor µ(x)=e^{\int -1/x dx}=e^{-\ln x}=x^{-1}.

Multiply whole equation by \(x^{-1}\): \(\dfrac{y+x}{x}dx + \ln x \, dy = 0 \Rightarrow \left(\dfrac{y}{x}+1\right)dx + \ln x \, dy = 0.\)

Now \(M = y/x +1 \Rightarrow M_y = 1/x.\) \(N = \ln x \Rightarrow N_x = 1/x.\) Exact.

Find potential \(\Phi\) with \(\Phi_x = y/x + 1\): \(\Phi = y\ln x + x + h(y).\)

\(\Phi_y = \ln x + h'(y) = N = \ln x \Rightarrow h'(y)=0 \Rightarrow h= C.\)

Solution: \(y\ln x + x = C.\)

Q31. Solve: \(\displaystyle \frac{dy}{dx} = \frac{x^2 - y^2}{xy}.\)

Homogeneous. Put \(y = vx \Rightarrow dy/dx = v + x dv/dx.\)

Substitute: \(v + x\frac{dv}{dx} = \dfrac{1 - v^2}{v}.\)

\(x\frac{dv}{dx} = \dfrac{1 - v^2}{v} - v = \dfrac{1 - 2v^2}{v}.\)

Separate: \(\dfrac{v}{1 - 2v^2} dv = \dfrac{dx}{x}.\)

Let \(u = 1 - 2v^2\), then \(du = -4v dv\). So \(\int \dfrac{v}{1 - 2v^2} dv = -\tfrac{1}{4}\ln|1 - 2v^2|.\)

Hence \(-\tfrac{1}{4}\ln|1 - 2v^2| = \ln|x| + C.\) So \(1 - 2\left(\dfrac{y}{x}\right)^2 = K x^{-4}.\)

Final implicit form: \( \boxed{1 - 2\frac{y^2}{x^2} = \dfrac{K}{x^4}}.\)

Q32. Solve: \(\displaystyle \frac{dy}{dx} = \frac{y + x e^{-x}}{x}.\)

Rewrite: \(\frac{dy}{dx} - \frac{y}{x} = e^{-x}.\) Linear with \(P(x)=-1/x,\ Q=e^{-x}\).

I.F. \(= e^{\int -1/x dx} = x^{-1}.\) Multiply: \(\dfrac{d}{dx}\left(\frac{y}{x}\right) = \frac{e^{-x}}{x}.\)

Integrate RHS: \(\int \frac{e^{-x}}{x} dx\) is the exponential integral Ei-type (no elementary primitive).

So solution in closed form is \( \dfrac{y}{x} = \int \frac{e^{-x}}{x} dx + C\), or \( y = x\left(C + \operatorname{Ei}(-x)\right)\) using the exponential integral notation.

Q33. Solve: \(\displaystyle \frac{dy}{dx} = \frac{x + y + 1}{x - y + 1}.\)

Use substitution to shift origin: let \(u = x+1,\ v = y\). Then equation becomes \(\dfrac{dv}{du} = \dfrac{u + v}{u - v}.\)

Now homogeneous in \(u,v\). Put \(v = w u\). Then \(dv/du = w + u dw/du\).

Substitute: \(w + u\frac{dw}{du} = \dfrac{1 + w}{1 - w}.\)

Solve: \(u\frac{dw}{du} = \dfrac{1 + w}{1 - w} - w = \dfrac{1 + w^2}{1 - w}.\)

Separate: \(\dfrac{1 - w}{1 + w^2} dw = \dfrac{du}{u}.\) Integrate and back-substitute \(w = \dfrac{y}{x+1}\) to get an implicit relation.

Q34. Solve: \(\displaystyle x\,dy - y\,dx = x^2 e^{x/y} \, dx.\)

Rearrange to differential in \(y/x\). Divide both sides by \(x^2\): \(\dfrac{dy}{x} - \dfrac{y}{x^2} dx = e^{x/y} dx.\)

This is nonstandard; better to use substitution \(v = \dfrac{y}{x}\Rightarrow y = vx,\ dy = v dx + x dv.\)

Substitute: \(x(v dx + x dv) - vx dx = x^2 e^{1/v} dx.\) Simplify left: \(x^2 dv = x^2 e^{1/v} dx.\)

So \(dv = e^{1/v} dx.\) Separate: \(dx = e^{-1/v} dv.\)

Integrate: \(x + C = \int e^{-1/v} dv\). The integral has no elementary primitive; state implicit solution: \( x = \int e^{-1/v} dv + C\) with \(v = y/x\).

Q35. Solve the linear equation: \(\displaystyle \frac{dy}{dx} + \frac{2x}{1+x^2} y = \sin x.\)

I.F. \(= e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2.\)

Multiply: \(\dfrac{d}{dx}\left(y(1+x^2)\right) = (1+x^2)\sin x.\)

Integrate RHS: \(\int (1+x^2)\sin x \, dx = \int \sin x dx + \int x^2 \sin x \, dx.\)

Use integration by parts for \(\int x^2 \sin x \, dx\). Final solution:

\(y(1+x^2) = -\cos x + ( -x^2\cos x + 2x\sin x + 2\cos x) + C\) (combine terms as needed). Then divide by \(1+x^2\).

Q36. Find particular solution of \(\displaystyle \frac{dy}{dx} = \frac{y}{x} + x e^{x^2}\) that satisfies \(y(0)=0.\)

Write as \(dy/dx - \dfrac{y}{x} = x e^{x^2}.\) I.F. \(= x^{-1}.\)

\(\dfrac{d}{dx}\left(\frac{y}{x}\right) = e^{x^2}.\) Integrate: \(\dfrac{y}{x} = \int e^{x^2} dx + C.\)

\(\int e^{x^2} dx\) is non-elementary (error function). Use initial condition \(y(0)=0\) carefully: divide form invalid at \(x=0\). Instead solve directly by substitution for small neighborhood or note that solution expressed as \(y = x\left(C + \int_0^x e^{t^2} dt\right)\).

Use \(y(0)=0\) ⇒ constant \(C=0\). Hence \( \boxed{y = x\int_0^x e^{t^2} dt }.\)

Q37. Solve: \(\displaystyle \frac{dy}{dx} = \frac{y^2 + 1}{x} - \frac{y}{x}.\)

Rewrite: \(\dfrac{dy}{dx} + \dfrac{y}{x} = \dfrac{y^2+1}{x}.\) Multiply by 2y to use substitution \(u=y^2\).

Let \(u=y^2\). Then \(du/dx = 2y dy/dx.\) Multiply equation by \(2y\):

\(du/dx + \dfrac{2u}{x} = \dfrac{2y(y^2+1)}{x}.\) Not linear in u due to RHS; alternate: rearrange original as \(dy/dx = \dfrac{y^2 - y + 1}{x}.\)

Use substitution \(y = vx\) not helpful. This is a Riccati-like with variable coefficients. State implicit solution method: look for particular solution or convert to linear if a particular solution known. For many exam problems one finds particular solution \(y=x\tan(\ln x + C)\) type after trial; otherwise solve numerically.

Q38. Solve the exact equation: \((2xy + y^2)\,dx + (x^2 + 2xy)\,dy = 0.\)

Check exactness: \(M=2xy+y^2 \Rightarrow M_y=2x+2y.\) \(N=x^2+2xy \Rightarrow N_x=2x+2y.\) Exact.

Find potential \(\Phi\) with \(\Phi_x = M\):

\(\Phi = \int (2xy+y^2) dx = x^2 y + x y^2 + h(y).\)

\(\Phi_y = x^2 + 2xy + h'(y) = N = x^2 + 2xy\Rightarrow h'(y)=0.\)

Solution: \( \boxed{x^2 y + x y^2 = C}.\)

Q39. Solve: \(\displaystyle \frac{dy}{dx} = \tan\left(\frac{y}{x}\right).\)

Put \(y = vx \Rightarrow dy/dx = v + x dv/dx.\)

Substitute: \(v + x\frac{dv}{dx} = \tan v.\)

So \(x\frac{dv}{dx} = \tan v - v.\) Separate: \(\dfrac{dv}{\tan v - v} = \dfrac{dx}{x}.\)

Integrate both sides: \(\int \dfrac{dv}{\tan v - v} = \ln|x| + C.\) The integral is non-elementary; leave implicit solution in this integral form and back-substitute \(v=y/x\).

Q40. Solve: \(\displaystyle \frac{dy}{dx} = \frac{2xy + x^3}{x^2}.\)

Simplify RHS: \(\dfrac{2xy}{x^2} + \dfrac{x^3}{x^2} = 2\frac{y}{x} + x.\)

Let \(y = vx \Rightarrow dy/dx = v + x dv/dx.\)

Substitute: \(v + x\frac{dv}{dx} = 2v + x.\) So \(x\frac{dv}{dx} = v + x.\)

Rearrange: \(\dfrac{dv}{dx} - \dfrac{v}{x} = 1.\) This is linear in v with I.F. \(= e^{\int -1/x dx} = x^{-1}.\)

\(\dfrac{d}{dx}\left(\frac{v}{x}\right) = \frac{1}{x}.\) Integrate: \(\frac{v}{x} = \ln|x| + C.\)

So \(v = x(\ln|x| + C)\) and \(y = vx = x^2(\ln|x| + C).\)

FAQ1. What is a differential equation?

A differential equation is an equation that relates a function with its derivatives. Example: \(\frac{dy}{dx} = 3x^2\).

FAQ2. What is the order and degree of a differential equation?

The order is the highest derivative in the equation. The degree is the power of the highest derivative, provided the equation is polynomial in derivatives. Example: \( (d^2y/dx^2)^3 + dy/dx = 0\) → Order = 2, Degree = 3.

FAQ3. What is a linear differential equation?

A linear differential equation has the function and its derivatives only to the first power and is not multiplied together. Example: \(dy/dx + P(x)y = Q(x)\).

FAQ4. What is a homogeneous differential equation?

A first-order DE \(\frac{dy}{dx} = F(y/x)\) is called homogeneous. It can be solved using substitution \(y = vx\).

FAQ5. How do you solve a first-order linear differential equation?

Write in standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), find the integrating factor \(IF = e^{\int P(x)dx}\), multiply through, then integrate: \(y \cdot IF = \int Q(x) IF dx + C\).

FAQ6. How do you identify a Bernoulli differential equation?

A Bernoulli DE is of the form \(\frac{dy}{dx} + P(x)y = Q(x)y^n\), with \(n \neq 0,1\). Solve by substitution \(v = y^{1-n}\) to reduce it to linear form.

FAQ7. What is an exact differential equation?

An equation \(M(x,y)dx + N(x,y)dy = 0\) is exact if \(M_y = N_x\). Solve by finding a potential function \(\Phi(x,y)\) such that \(d\Phi = Mdx + Ndy\).

FAQ8. How do you find an integrating factor?

If a DE is not exact, an integrating factor \(\mu(x)\) or \(\mu(y)\) may make it exact. For example, \(\mu(x) = e^{\int (M_y - N_x)/N dx}\) if the expression depends only on x.

FAQ9. Can all first-order DEs be solved analytically?

No, some first-order DEs cannot be solved in terms of elementary functions. Examples include \(\frac{dy}{dx} = e^{x^2}\) or \(\frac{dy}{dx} = \tan(y/x)\). Such solutions may use special functions or numerical methods.

FAQ10. What is the difference between particular and general solution?

The general solution contains all possible solutions with arbitrary constants. A particular solution is obtained by applying initial/boundary conditions to fix the constants. Example: \(y = Ce^x\) is general; \(y = 2e^x\) is particular if \(y(0)=2\).

FAQ11. What is a Riccati differential equation?

A Riccati equation has the form \(\frac{dy}{dx} = P(x) + Q(x)y + R(x)y^2\). It is nonlinear but can sometimes be reduced to a linear equation if a particular solution is known.

FAQ12. How do you solve a homogeneous linear DE with constant coefficients?

For an nth-order DE like \(a_n y^{(n)} + ... + a_1 y' + a_0 y = 0\), form the characteristic equation \(a_n r^n + ... + a_1 r + a_0 = 0\). Solve for roots and write the solution based on real, repeated, or complex roots.

FAQ13. What is meant by separable differential equation?

A first-order DE is separable if it can be written as \(N(y)dy = M(x)dx\). Integrate both sides to find the solution.

FAQ14. How do you solve a DE using substitution \(y = vx\)?

When the DE is homogeneous (\(\frac{dy}{dx} = F(y/x)\)), put \(y = vx\). Then \(dy/dx = v + x dv/dx\). Substitute and separate variables in terms of \(v\) and \(x\) to integrate.

FAQ15. What are exact and inexact equations?

An equation \(Mdx + Ndy = 0\) is exact if \(M_y = N_x\). Otherwise, it’s inexact. An integrating factor can sometimes make an inexact equation exact.

FAQ16. Can a DE have more than one particular solution?

No, a particular solution is unique for a given initial/boundary condition. Changing the condition gives a different particular solution.

FAQ17. How do you solve a Bernoulli DE when \(n=2\)?

For \(\frac{dy}{dx} + P(x)y = Q(x)y^2\), put \(v = 1/y\). Then \(dv/dx - P(x)v = -Q(x)\). Solve this linear DE in \(v\), then revert \(y = 1/v\).

FAQ18. What is a first-order linear DE with variable coefficients?

It has the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x)\) and \(Q(x)\) are functions of \(x\). Solve using integrating factor \(IF = e^{\int P(x) dx}\).

FAQ19. How do you verify a solution of a DE?

Substitute the solution into the original differential equation. If both sides match, the solution is correct.

FAQ20. How do you approach higher-order DEs?

For higher-order linear DEs, check if coefficients are constant or variable. Use characteristic equation for constant coefficients. For variable coefficients, try reduction of order, known substitutions, or special methods depending on the form.

FAQ21. What is a partial differential equation (PDE)?

A PDE involves partial derivatives of a function with more than one independent variable. Example: \(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0\).

FAQ22. How do you reduce a Riccati equation to a linear DE?

If a particular solution \(y_p\) is known, put \(y = y_p + 1/v\). This substitution transforms the Riccati equation into a linear first-order DE in \(v\).

FAQ23. What is the difference between linear and nonlinear DEs?

Linear DEs involve the function and its derivatives only to the first power, with no products between them. Nonlinear DEs include powers, products, or functions like sin(y) or e^y.

FAQ24. How do you solve an exact DE if it becomes inexact after simplification?

Use an integrating factor depending on \(x\) or \(y\) to convert it back to exact. Then find the potential function and write the solution.

FAQ25. What is a first-order linear DE with constant coefficients?

It has the form \(\frac{dy}{dx} + ay = b\), where \(a, b\) are constants. Solve using \(y = Ce^{-ax} + \frac{b}{a}\) if \(a \neq 0\).

FAQ26. How do you solve a DE using separation of variables?

Rewrite the DE as \(N(y)dy = M(x)dx\), then integrate both sides. Solve for \(y\) to get the solution. Works only if variables can be separated.

FAQ27. What is a second-order linear DE?

A DE of the form \(a(x)y'' + b(x)y' + c(x)y = f(x)\). Solve using characteristic equation if coefficients are constant or reduction of order / method of undetermined coefficients if variable.

FAQ28. How do you identify an exact equation quickly?

Compute \(M_y\) and \(N_x\). If \(M_y = N_x\), the equation is exact. Otherwise, consider an integrating factor.

FAQ29. How do you solve a first-order linear DE with variable coefficients?

Standard form: \(dy/dx + P(x)y = Q(x)\). Use integrating factor \(IF = e^{\int P(x) dx}\), then multiply through: \(d/dx(y \cdot IF) = Q(x) IF\). Integrate RHS and solve for \(y\).

FAQ30. Can solutions of differential equations be implicit?

Yes, sometimes the solution is expressed as an implicit relation between \(x\) and \(y\), like \(x^2 + y^2 = C\). It’s valid as long as it satisfies the differential equation.

FAQ31. What is a linear second-order DE with constant coefficients?

It has the form \(ay'' + by' + cy = f(x)\) with constants \(a, b, c\). Solve the homogeneous part using the characteristic equation \(ar^2 + br + c = 0\), then find a particular solution for \(f(x)\).

FAQ32. How do you solve a DE using substitution \(y = ux + v\)?

This substitution is useful when DE involves combinations like \(y - x\) or \(y + x\). Replace \(y\) and \(dy/dx\) in terms of \(u, v\) to reduce the DE to separable or linear form.

FAQ33. What is an integrating factor and how do you find it?

An integrating factor (IF) is a function that, when multiplied with a non-exact DE, makes it exact. Often \(\mu(x) = e^{\int (M_y - N_x)/N dx}\) if it depends on \(x\), or similarly for \(y\).

FAQ34. What are singular solutions of a DE?

Singular solutions are solutions that cannot be obtained by choosing constants in the general solution. They often satisfy the differential equation but not the form of the general solution.

FAQ35. How do you solve a homogeneous linear DE with repeated roots?

If the characteristic equation has repeated root \(r\), then the solution is \(y = (C_1 + C_2 x)e^{rx}\) for second-order, extend similarly for higher-order DEs.

FAQ36. How do you solve a DE of the form \(dy/dx = f(ax + by + c)\)?

Use the substitution \(u = ax + by + c\). Then \(du/dx = a + b dy/dx\). Substitute dy/dx in terms of u and separate variables to integrate.

FAQ37. What is the method of variation of parameters?

Used to find a particular solution of non-homogeneous linear DEs. Assume constants in the homogeneous solution are functions of \(x\) and solve for them to satisfy the DE.

FAQ38. How do you handle DEs that cannot be solved by elementary functions?

Leave the solution in integral form or use special functions like error function (erf), exponential integral (Ei), or apply numerical methods if required.

FAQ39. How do you check if a DE is exact after multiplying by an integrating factor?

Compute \(M_y\) and \(N_x\) after multiplying by the integrating factor. If they are equal, the equation is exact, and you can solve it using the potential function method.

FAQ40. What is the general approach to solving first-order non-linear DEs?

Try to identify type: separable, homogeneous, exact, Bernoulli, or Riccati. Use appropriate substitution, integrating factor, or transformation to reduce it to a linear or integrable form.

FAQ41. What is a complementary function in DEs?

The complementary function (CF) is the general solution of the homogeneous part of a linear DE. It represents all possible solutions without the forcing term.

FAQ42. What is a particular integral in DEs?

The particular integral (PI) is a specific solution of the non-homogeneous DE that satisfies the full equation, including the forcing term. The general solution is CF + PI.

FAQ43. How do you solve \(\frac{dy}{dx} = y^2 + x^2\)?

This is a Riccati-type equation with no simple elementary solution. Solution may require substitution if a particular solution is known or leave in implicit/integral form for higher-level exams.

FAQ44. How do you find singular solutions for Clairaut’s equation?

For Clairaut’s equation \(y = xy' + f(y')\), the singular solution is obtained by eliminating \(y'\) between \(y = xy' + f(y')\) and \(\frac{dy}{dy'} = 0\).

FAQ45. How do you reduce a second-order DE to first order?

If DE is of the form \(y'' = f(x,y')\), set \(p = y'\). Then \(dp/dx = f(x,p)\). Solve the first-order DE for \(p\), then integrate \(p = dy/dx\) to get \(y\).

FAQ46. How do you handle DEs with variable separable coefficients?

Rewrite the DE so all \(y\) terms are on one side and \(x\) terms on the other. Integrate each side, then solve for \(y\). Example: \(\frac{dy}{dx} = g(x)h(y)\).

FAQ47. How do you solve a DE using the substitution \(y^n = v\)?

This is the standard method for Bernoulli DE: \(\frac{dy}{dx} + P(x)y = Q(x)y^n\). Let \(v = y^{1-n}\) to reduce to linear DE in \(v\).

FAQ48. What are the common methods to solve non-homogeneous linear DEs?

Common methods: 1) Method of undetermined coefficients, 2) Variation of parameters, 3) Using Laplace transforms (for constant coefficients or special cases).

FAQ49. How do you deal with DEs with trigonometric or exponential forcing functions?

Use the method of undetermined coefficients. Guess a particular solution similar in form to the forcing function (sin, cos, e^x) and solve for constants.

FAQ50. How do you verify solutions for advanced DEs?

Substitute the proposed solution into the original DE. Simplify carefully; for non-elementary integrals or implicit forms, differentiate or check consistency with initial/boundary conditions.

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