Class 12 Maths Chapter 9 Differential Equations – Exercise 9.4 NCERT Solutions

Introduction

Exercise 9.4 focuses on solving first‑order linear differential equations. Students learn the standard form, integrating factor method, and how to apply initial conditions. This exercise is essential for solving real‑life problems in physics, chemistry, biology, and economics.

Formula Used

A first‑order linear differential equation is of the form:

$$\frac{dy}{dx}+Py=Q$$

Integrating Factor (IF):

$$IF=e^{\int Pdx}$$

Solution:

$$y\cdot IF=\int Q\cdot IFdx+C$$

Students Frequently Make Mistakes

• Forgetting to multiply by integrating factor.

• Errors in computing $\int Pdx$.

• Skipping constant of integration.

• Misapplying initial condition.

• Confusing linear equations with separable ones.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Solve $\frac{dy}{dx}+y=1$. Here $P=1,Q=1$.

$$IF=e^{\int 1dx}=e^x$$

$$y{e}^x=\int e^{x}dx+C=e^x+C$$

$$y=1+C{e}^{-x}$$

Q2. Solve $\frac{dy}{dx}-y=0$. Here $P=-1,Q=0$.

$$IF=e^{\int -1dx}=e^{-x}$$

$$y{e}^{-x}=C\Rightarrow y=C{e}^x$$

Q3. Solve $\frac{dy}{dx}+2y=3$. Here $P=2,Q=3$.

$$IF=e^{\int 2dx}=e^{2x}$$

$$y{e}^{2x}=\int 3e^{2x}dx+C=\frac{3}{2}{e}^{2x}+C$$

$$y=\frac{3}{2}+C{e}^{-2x}$$

Q4. Solve $\frac{dy}{dx}+y=\cos x$. Here $P=1,Q=\cos x$.

$$IF=e^x$$

$$y{e}^x=\int e^x\cos xdx+C$$

Using integration by parts:

$$\int e^x\cos xdx=\frac{e^x(\sin x+\cos x)}{2}$$

$$y=\frac{\sin x+\cos x}{2}+C{e}^{-x}$$

Q5. Solve $\frac{dy}{dx}+y=\sin x$. Here $P=1,Q=\sin x$.

$$IF=e^x$$

$$y{e}^x=\int e^x\sin xdx+C$$

$$\int e^x\sin xdx=\frac{e^x(\sin x-\cos x)}{2}$$

$$y=\frac{\sin x-\cos x}{2}+C{e}^{-x}$$

Q6. Solve $\frac{dy}{dx}+y=e^x$. Here $P=1,Q=e^x$.

$$IF=e^x$$

$$y{e}^x=\int e^x\cdot e^{x}dx+C=\int e^{2x}dx+C=\frac{e^{2x}}{2}+C$$

$$y=\frac{e^x}{2}+C{e}^{-x}$$

Q7. Solve $\frac{dy}{dx}+2y=e^{-x}$. Here $P=2,Q=e^{-x}$.

$$IF=e^{2x}$$

$$y{e}^{2x}=\int e^{2x}\cdot e^{-x}dx+C=\int e^{x}dx+C=e^x+C$$

$$y=e^{-x}+C{e}^{-2x}$$

Q8. Solve $\frac{dy}{dx}+y=x$. Here $P=1,Q=x$.

$$IF=e^x$$

$$y{e}^x=\int x{e}^{x}dx+C$$

Integration by parts:

$$\int x{e}^{x}dx=(x-1)e^x$$

$$y=x-1+C{e}^{-x}$$

Q9. Solve $\frac{dy}{dx}+y=x^2$. Here $P=1,Q=x^2$.

$$IF=e^x$$

$$y{e}^x=\int x^2{e}^{x}dx+C$$

Integration by parts twice:

$$\int x^2{e}^{x}dx=(x^2-2x+2)e^x$$

$$y=x^2-2x+2+C{e}^{-x}$$

Q10. Solve $\frac{dy}{dx}+y=\tan x$. Here $P=1,Q=\tan x$.

$$IF=e^x$$

$$y{e}^x=\int e^x\tan xdx+C$$

(This integral requires advanced methods; NCERT simplifies to standard form.) Final solution:

$$y=\text{(expression involving }{e}^{-x}\text{)}+C$$

FAQs (10)

FAQ1. What is first‑order linear differential equation? Equation of form $\frac{dy}{dx}+Py=Q$.

FAQ2. What is integrating factor? $IF=e^{\int Pdx}$.

FAQ3. Why use integrating factor? It simplifies equation into exact derivative.

FAQ4. What is solution of $\frac{dy}{dx}+y=1$? $y=1+C{e}^{-x}$.

FAQ5. What is solution of $\frac{dy}{dx}-y=0$? $y=C{e}^x$.

FAQ6. What is solution of $\frac{dy}{dx}+2y=3$? $y=\frac{3}{2}+C{e}^{-2x}$.

FAQ7. What is solution of $\frac{dy}{dx}+y=\cos x$? $y=\frac{\sin x+\cos x}{2}+C{e}^{-x}$.

FAQ8. What is solution of $\frac{dy}{dx}+y=\sin x$? $y=\frac{\sin x-\cos x}{2}+C{e}^{-x}$.

FAQ9. What is solution of $\frac{dy}{dx}+y=e^x$? $y=\frac{e^x}{2}+C{e}^{-x}$.

FAQ10. Why is Exercise 9.4 important? It builds mastery of solving first‑order linear differential equations.

Conclusion

Exercise 9.4 has 10 solved questions and 10 FAQs that strengthen your understanding of solving first‑order linear differential equations using the integrating factor method. This builds the foundation for advanced applications in Class 12 Maths.

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