Class 12 Maths Chapter 9 Differential Equations – Exercise 9.3 NCERT Solutions

Introduction

Exercise 9.3 focuses on solving differential equations by the method of separation of variables. Students learn how to rearrange equations so that variables can be separated and integrated independently. This technique is widely used in physics, chemistry, biology, and economics.

Formula Used

If $\frac{dy}{dx}=f(x)g(y)$, then

$$\frac{dy}{g(y)}=f(x)dx$$

Integrate both sides:

$$\int \frac{1}{g(y)}dy=\int f(x)dx$$

Students Frequently Make Mistakes

• Forgetting to separate variables correctly.

• Errors in integration of trigonometric/exponential functions.

• Skipping constant of integration.

• Misapplying logarithmic rules.

• Confusing general and particular solutions.

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

Q1. Solve $\frac{dy}{dx}=x+y$. Not directly separable. Rewrite: $\frac{dy}{dx}-y=x$. This is linear, solved later. (Here, Exercise 9.3 focuses on separable ones.)

Q2. Solve $\frac{dy}{dx}=xy$.

$$\frac{dy}{y}=xdx$$

$$\ln \mathrm{\mid }y\mathrm{\mid }=\frac{x^2}{2}+C\Rightarrow y=C{e}^{x^2\mathrm{/}2}$$

Q3. Solve $\frac{dy}{dx}=y$.

$$\frac{dy}{y}=dx$$

$$\ln \mathrm{\mid }y\mathrm{\mid }=x+C\Rightarrow y=C{e}^x$$

Q4. Solve $\frac{dy}{dx}=x$.

$$dy=xdx$$

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

Q5. Solve $\frac{dy}{dx}=y^2$.

$$\frac{dy}{y^2}=dx$$

$$-\frac{1}{y}=x+C\Rightarrow y=\frac{-1}{x+C}$$

Q6. Solve $\frac{dy}{dx}=x^{2}y$.

$$\frac{dy}{y}=x^{2}dx$$

$$\ln \mathrm{\mid }y\mathrm{\mid }=\frac{x^3}{3}+C\Rightarrow y=C{e}^{x^3\mathrm{/}3}$$

Q7. Solve $\frac{dy}{dx}=y\cos x$.

$$\frac{dy}{y}=\cos xdx$$

$$\ln \mathrm{\mid }y\mathrm{\mid }=\sin x+C\Rightarrow y=C{e}^{\sin x}$$

Q8. Solve $\frac{dy}{dx}=y\sin x$.

$$\frac{dy}{y}=\sin xdx$$

$$\ln \mathrm{\mid }y\mathrm{\mid }=-\cos x+C\Rightarrow y=C{e}^{-\cos x}$$

Q9. Solve $\frac{dy}{dx}=y{e}^x$.

$$\frac{dy}{y}=e^{x}dx$$

$$\ln \mathrm{\mid }y\mathrm{\mid }=e^x+C\Rightarrow y=C{e}^{e^x}$$

Q10. Solve $\frac{dy}{dx}=y\tan x$.

$$\frac{dy}{y}=\tan xdx$$

$$\ln \mathrm{\mid }y\mathrm{\mid }=-\ln \mathrm{\mid }\cos x\mathrm{\mid }+C\Rightarrow y=C\sec x$$

FAQs (10)

FAQ1. What is separation of variables? Method to solve differential equations by separating $x$ and $y$.

FAQ2. When can separation be applied? When equation can be written as $\frac{dy}{dx}=f(x)g(y)$.

FAQ3. Why add constant of integration? Because indefinite integrals represent families of solutions.

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

FAQ5. What is solution of $\frac{dy}{dx}=xy$? $y=C{e}^{x^2\mathrm{/}2}$.

FAQ6. What is solution of $\frac{dy}{dx}=y^2$? $y=\frac{-1}{x+C}$.

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

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

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

FAQ10. Why is Exercise 9.3 important? It builds mastery of solving separable differential equations.

Conclusion

Exercise 9.3 has 10 solved questions and 10 FAQs that strengthen your understanding of solving differential equations by separation of variables. This builds the foundation for advanced techniques in Class 12 Maths.

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