Differential Equation Previous Year Questions with Solutions

1.0Introduction

Differential Equations Previous Year Questions typically cover fundamental concepts like order and degree of a differential equation, solving first-order and first-degree equations (separable, linear, and homogeneous), and higher-order linear differential equations with constant coefficients. Common problems include solving equations using methods such as separation of variables, integrating factors, and applying initial or boundary conditions. Other types involve forming differential equations from given conditions or applying them to real-world scenarios like growth and decay or motion problems.

Solutions generally involve step-by-step application of solving techniques, integration, and algebraic manipulation. Practicing these previous year questions enhances conceptual clarity and problem-solving speed, which is especially helpful in competitive exams and board preparations.

2.0Differential Equation Previous Year Questions with Solutions

JEE questions in Differential Equations often assess understanding of the formation, order, and degree of differential equations, as well as methods for solving them. The focus is typically on first-order and first-degree equations, but questions can also involve second-order linear differential equations.

Some common types of problems include:

Formation of Differential Equations: Deriving a differential equation from a given family of curves by eliminating arbitrary constants.

First-Order Differential Equations: Solving using methods such as:

• Separation of Variables: Often used in problems involving growth/decay or cooling.
• Homogeneous Equations: Involving substitution like y = vx.
• Linear Differential Equations: Of the form $\frac{dy}{dx}+Py=Q$ , solved using integrating factors.

Second-Order Linear Differential Equations: May appear in advanced problems, especially in JEE Advanced, and involve constant coefficients or applications in motion and oscillation.

These questions are designed to test conceptual clarity, application skills, and familiarity with standard solving techniques.

Note: In the JEE Main Mathematics exam, you can generally expect 1 to 2 questions from the Differential Equations chapter, while JEE Advanced may feature more challenging or application-based problems.

3.0Key Concepts to Remember – Differential Equations

Order and Degree:

• Order: The highest order derivative in the equation.
• Degree: The highest power (exponent) of the highest order derivative, provided the equation is polynomial in derivatives.

Formation of Differential Equations:

• Formed by eliminating arbitrary constants from a given relation.
• Number of arbitrary constants = Order of the resulting differential equation.

Types of First-Order Differential Equations:

Separable Equations: Can be written as $\frac{dy}{dx}=f(x)g(y)$

Integrate both sides after separating variables.

Homogeneous Equations: Of the form $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$

Use substitution y = vx.

Linear Equations: Standard form $\frac{dy}{dx}+Py=Q$

Use Integrating Factor (IF): $IF=e^{\int Pdx}$

General solution: $y\cdot IF=\int Q\cdot IFdx+C$

Particular and General Solutions:

General Solution: Contains arbitrary constant(s).

Particular Solution: Obtained by applying initial conditions to the general solution.

Applications:

Problems involving growth/decay, cooling, mixing, and motion are based on first-order differential equations.

Often use exponential functions in their solutions.

Second-Order Linear Differential Equations (JEE Advanced level):

Typically of the form: $a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0$

Solution depends on the nature of the roots of the auxiliary equation.

4.0JEE Mains Past Year Questions with Solutions on Differential Equations 

1. Let y = y(x) be the solution of the differential equation (x² + 4)²dy + (2x³y + 8xy – 2)dx = 0. If y(0) = 0, then y(2) is equal to  

$$\begin{gathered} (1)\frac{\pi }{8} \\ (2)\frac{\pi }{16} \\ (3)2\pi \\ (4)\frac{\pi }{32} \end{gathered}$$

Ans. (4)

Sol.

$$\begin{gathered} \frac{dy}{dx}+y\left(\frac{2x^3+8x}{(x^2+4{)}^2}\right)=\frac{2}{(x^2+4{)}^2} \\ \frac{dy}{dx}+y\left(\frac{2x}{x^2+4}\right)=\frac{2}{(x^2+4{)}^2} \\ \text{IF}=e^{\int \frac{2x}{x^2+4}dx} \\ \text{IF}=x^2+4 \\ y\times (x^2+4)=\int \frac{2}{(x^2+4{)}^2}\cdot (x^2+4)dx \\ y(x^2+4)=2\int \frac{dx}{x^2+2^2} \\ y(x^2+4)=\frac{2}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C \\ y(x^2+4)={\tan }^{-1}\left(\frac{x}{2}\right) \\ y\text{ at }x=2 \\ y(4+4)={\tan }^{-1}(1) \\ \boxed{y(2)=\frac{\pi }{32}} \end{gathered}$$
Option (4) is correct

2. Let y = y(x) be the solution of the differential equation (x + y + 2)² dx = dy, y(0) = –2. {(x + y + 2)^2 dx = dy, \quad y(0) = -2} Let the maximum and minimum values of the function y = y(x) in $y=y(x)\text{ in }\left[0,\frac{\pi }{3}\right]$ be α and β, respectively. If , then γ + δ equals …..$(3\alpha +\pi {)}^2+{\beta }^2=\gamma +\delta \sqrt{3},\gamma ,\delta \in \mathbb{Z},\text{ then }\gamma +\delta \text{ equals}\dots$

Ans. (31)

Sol.

$$\begin{gathered} \frac{dy}{dx}=(x+y+2{)}^2....(1),y(0)=-2 \\ (x+y+2=v) \\ 1+\frac{dy}{dx}=\frac{dv}{dx} \\ \frac{dv}{dx}=1+v^2 \\ \int \frac{dv}{1+v^2}=\int dx \\ {\tan }^{-1}(v)=x+C \\ {\tan }^{-1}(x+y+2)=x+C \\ \Rightarrow {\tan }^{-1}(x+y+2)=x \\ y=\tan x-x-2 \\ f(x)=\tan x-x-2,x\in \left[0,\frac{\pi }{3}\right] \\ {f}^{\prime }(x)={\sec }^{2}x-1>0\Rightarrow f(x)\uparrow \\ {f}_{\min }=f(0)=-2=\beta \\ {f}_{\max }=f\left(\frac{\pi }{3}\right)=\sqrt{3}-\frac{\pi }{3}-2=\alpha \\ \text{Now }(3\alpha +\pi {)}^2+{\beta }^2=\gamma +\delta \sqrt{3} \\ \Rightarrow (3\alpha +\pi {)}^2+{\beta }^2={\left(3\left(\sqrt{3}-\frac{\pi }{3}-2\right)+\pi \right)}^2+4 \\ \Rightarrow \gamma +\delta \sqrt{3}=67-36\sqrt{3} \\ \Rightarrow \gamma =67\text{ and }\delta =-36\Rightarrow \gamma +\delta =31 \end{gathered}$$

3. If the solution y = y(x) of the differential equation $(x^4+2x^3+3x^2+2x+2)dy-(2x^2+2x+3)dx=0\text{satisfies }y(-1)=-\frac{\pi }{4},theny(0)$ is equal to:

$$\begin{gathered} (1)\frac{-}{\pi }12 \\ (2)0 \\ (3)\frac{}{\pi }4 \\ (4)\frac{\pi }{2} \end{gathered}$$

Ans. (3)

Sol.

$$\begin{gathered} \int dy=\int \frac{(2x^2+2x+3)}{x^4+2x^3+3x^2+2x+2}dx \\ y=\int \frac{(2x^2+2x+3)}{(x^2+1)(x^2+2x+2)}dx \\ y=\int \frac{dx}{x^2+2x+2}+\int \frac{dx}{x^2+1} \\ y={\tan }^{-1}(x+1)+{\tan }^{-1}x+C \\ y(-1)=-\frac{\pi }{4} \\ -\frac{\pi }{4}=0-\frac{\pi }{4}+C\Rightarrow C=0 \\ \Rightarrow y={\tan }^{-1}(x+1)+{\tan }^{-1}x \\ y(0)={\tan }^{-1}1=\frac{\pi }{4} \end{gathered}$$

4. Let y = y(x) be the solution of the differential equation $\frac{dy}{dx}=\frac{2x}{(1+x^2)}=y=x{e}^{\frac{1}{1+x^2}},y(0)=0$. Then the area enclosed by the curve $f(x)=y(x)e^{\frac{1}{1+x^2}}$ and the line y – x = 4 is ____.

Ans. (18)

$$\begin{gathered} \text{Sol.}IF=e^{\int \frac{2x}{(1+x^2{)}^2}dx}=e^{-\frac{1}{1+x^2}} \\ y\cdot e^{\frac{1}{1+x^2}}=\int x\cdot e^{\frac{1}{1+x^2}}\cdot e^{-\frac{1}{1+x^2}}dx \\ y\cdot e^{\frac{1}{1+x^2}}=\frac{x^2}{2}+C \\ (0,0)\Rightarrow \boxed{C=0} \\ y(x)=\frac{x^2}{2}\cdot e^{-\frac{1}{1+x^2}} \\ f(x)=\frac{x^2}{2} \\ A={\int }_{-2}^4(x+4)-\frac{x^2}{2}dx=18 \end{gathered}$$

5. Suppose the solution of the differential equation$\frac{dy}{dx}=\frac{(2+\alpha )x-\beta y+2}{\beta x-2\alpha y-(\beta y-4\alpha )}$ represents a circle passing through origin. Then the radius of this circle is :

$$\begin{gathered} (1)\sqrt{17} \\ (2)\frac{1}{2} \\ (3)\frac{\sqrt{17}}{2} \\ (4)2 \end{gathered}$$

Ans. (3)

Sol.

$$\begin{gathered} \frac{dy}{dx}=\frac{(2+\alpha )x-\beta y+2}{\beta x-2\alpha y-(\beta y-4\alpha )} \\ \beta xdy-(2\alpha +\beta )ydx+4\alpha dy=(2+\alpha )xdx-\beta ydx+2dx \\ \beta (xdy+ydx)-(2\alpha +\beta )ydx+4\alpha dy=(2+\alpha )xdx-\beta ydx+2dx \\ \beta xy-\frac{(2\alpha +\beta )y^2}{2}+4\alpha y=\frac{(2+\alpha )x^2}{2} \\ \Rightarrow \beta =0\text{for this to be a circle} \\ (2+\alpha )x^2+\alpha y^2+2x-4\alpha y=0 \\ \text{coeff. of }{x}^2=y^2\Rightarrow 2+\alpha =2\alpha \Rightarrow \alpha =2 \\ \Rightarrow \alpha =2 \\ 2x^2+2y^2+2x-8y=0 \\ {x}^2+y^2+x-4y=0 \\ \text{rd}=\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2} \end{gathered}$$

6. If  $x^3\sin \left(\frac{1}{x}\right),x\ne 0$ ,  x = 0

$$\begin{gathered} (1)(f^{\prime \prime }(0)=1) \\ (2)(f^{\prime \prime }\left(\frac{2}{\pi }\right)=\frac{24-{\pi }^2}{2\pi }) \\ (3)(f^{\prime \prime }\left(\frac{2}{\pi }\right)=\frac{12-{\pi }^2}{2\pi }) \\ (4)(f^{\prime \prime }(0)=0) \end{gathered}$$

Ans. (2)

Sol.

$$\begin{gathered} f^{\prime }(x)=3x^2\sin \left(\frac{1}{x}\right)-x\cos \left(\frac{1}{x}\right) \\ {f}^{\prime \prime }(x)=6x\sin \left(\frac{1}{x}\right)-3\cos \left(\frac{1}{x}\right)+\cos \left(\frac{1}{x}\right)-\sin \left(\frac{1}{x}\right) \\ {f}^{\prime \prime }\left(\frac{2}{\pi }\right)=\frac{12}{\pi }-\frac{\pi }{2}=\frac{24-{\pi }^2}{2\pi } \end{gathered}$$

7. Let y = y(x) be the solution of the differential equation $(2x\log x)\frac{dy}{dx}+2y=\frac{3}{x}\log x,x>0,y(e^1)=0$ . Then, y(e) is equal to 

$$\begin{gathered} (1)\frac{-}{3}2e \\ (2)\frac{-}{2}3e \\ (3)\frac{-}{3}e \\ (4)\frac{-}{2}e \end{gathered}$$

Ans. (3)

Sol.

$$\begin{gathered} \frac{dy}{dx}+\frac{y}{x\ln x}=\frac{3}{2x^2} \\ \therefore \text{I.F.}=e^{\int \frac{1}{x\ln x}dx} \\ \therefore y\ln x=\int \frac{3\ln x}{2x^2}dx \\ =\frac{3\ln x}{2}\int x^{-2}dx-\int \left(\frac{3}{2x}\int x^{-2}dx\right)dx \\ =\frac{3\ln x}{2x}-\int \frac{3}{2x}\cdot \left(\int x^{-2}dx\right)dx \\ y\cdot \square \cdot \ln x=\frac{-3\ln x}{2x}-\frac{3}{2x}+C \\ y(e^{-1})=0 \\ \Rightarrow 0=\frac{-3e}{2}+\frac{3e}{2}+C\Rightarrow C=0 \\ y=\frac{-3\ln x}{2x}+\frac{3}{2x} \\ \therefore y(e)=\frac{-3}{2e}+\frac{3}{2e}=\frac{-3+3}{2e}=0 \\ \boxed{y(e)=\frac{-3}{2e}+\frac{3}{2e}=\frac{-3+3}{2e}=0} \end{gathered}$$

8. Let y = y(x) be the solution curve of the differential equation secy $\frac{dy}{dx}+2x\sin y=x^2\cos y,y(1)=0\text{. Then }y(\sqrt{3})$ is equal to :

$$\begin{gathered} (1)\frac{\pi }{3} \\ (2)\frac{\pi }{6} \\ (3)\frac{}{\pi }4 \\ (4)\frac{\pi }{12} \end{gathered}$$

Ans. (3)

Sol.

$$\begin{gathered} \frac{dy}{dx}\cdot {\sec }^{2}y+2x\sin y\cdot \sec y=x^2\cos y\cdot \sec y \\ {\sec }^{2}y\cdot \frac{dy}{dx}+2x\tan y=x^3 \\ (\tan y=t)， \\ \frac{dt}{dx}+2xt=x^3 \\ (\text{IF}=e^{\int 2xdx}=e^{x^2}) \\ t{e}^{x^2}=\int x^3{e}^{x^2}dx+c \\ (x^2=Z) \\ t{e}^Z=\frac{1}{2}\int e^{Z}ZdZ=\frac{1}{2}[e^{Z}Z-\int e^{Z}dZ]+c \\ 2\tan y=(x^2-1)+2c{e}^{-x^2} \\ y(1)=0\Rightarrow c=0\Rightarrow y(\sqrt{3})=\frac{\pi }{4} \end{gathered}$$

9. Let , ${\int }_0^x\sqrt{1-(y^{\prime }(t){)}^2}dt={\int }_0^{x}y(t)dt,0\le x\le 3,y\ge 0,y(0)=0$. Then at $(x=2),thevalueof(y^{\prime \prime }+y^{\prime }+1)$ is equal to :

$$\begin{gathered} (1)1 \\ (2)2 \\ (3)\sqrt{2} \\ (4)\frac{1}{2} \end{gathered}$$

Ans. (1)

Sol.

$$\begin{gathered} \sqrt{1-(y^{\prime }(x){)}^2}=y(x) \\ {\left(\frac{dy}{dx}\right)}^2=1-y^2 \\ \frac{dy}{\sqrt{1-y^2}}=dx\text{or}\frac{dy}{\sqrt{1-y^2}}=-dx \\ \Rightarrow {\sin }^{-1}y=x+c\Rightarrow {\sin }^{-1}y=-x+c \\ x=0,y=0\Rightarrow c=0 \\ {\sin }^{-1}y=x,\text{as }y\ge 0 \\ \Rightarrow \sin x=y \\ \frac{dy}{dx}=\cos x \\ \frac{d^{2}y}{d{x}^2}=-\sin x \\ \Rightarrow -\sin x+\sin x+1=1 \end{gathered}$$

10. The solution curve, of the differential equation $2y\frac{dy}{dx}+3=5\frac{dy}{dx}$, passing through the point (0, 1) is a conic, whose vertex lies on the line :

(1) 2x + 3y = 9

(2) 2x + 3y = –9

(3) 2x + 3y = –6

(4) 2x + 3y = 6

Ans. (1)

Sol.

$$\begin{gathered} (2y-5)\frac{dy}{dx}=-3 \\ (2y-5)dy=-3dx \\ \int (2y-5)dy=\int -3dx \\ {y}^2-5y=-3x+\lambda \Rightarrow \frac{2y^2}{2}-5y=-3x+\lambda \\ \Rightarrow \lambda =-4 \\ {\left(y-\frac{5}{2}\right)}^2=-3\left(x-\frac{3}{4}\right) \\ \therefore \text{Vertex of parabola will be }\left(\frac{3}{4},\frac{5}{2}\right) \end{gathered}$$

11. The solution of the differential equation $(x^2+y^2)dx-5xydy=0,y(1)=0$, is :

$$\begin{gathered} (1)|x^2-4y^2{|}^5=x^2 \\ (2)|x^2-2y^2{|}^5=x \\ (3)|x^2-4y^2{|}^6=x \\ (4)|x^2-2y^2{|}^5=x^2 \end{gathered}$$

Ans. (1)

Sol.

$$\begin{gathered} (x^2+y^2)dx=5xy,dy \\ \Rightarrow \frac{dy}{dx}=\frac{x^2+y^2}{5xy} \\ (y=Vx) \\ \Rightarrow V+x\frac{dV}{dx}=\frac{1+V^2}{5V} \\ x\frac{dV}{dx}=\frac{1-4V^2}{5V} \\ \Rightarrow \int \frac{V}{1-4V^2},dV=\int \frac{dx}{5x} \\ (1-4V^2=t)， \\ -8V,dV=dt \\ \Rightarrow \int \frac{dt}{(-8)t}=\int \frac{dx}{5x} \\ -\frac{1}{8}\ln |t|=\frac{1}{5}\ln |x|+\ln C \\ \Rightarrow -5\ln |t|=8\ln |x|+\ln K \\ \Rightarrow \ln x^8+\ln |t{|}^5+\ln K=0 \\ \Rightarrow x^8|t{|}^5=C \\ \Rightarrow x^8{\left|1-4V^2\right|}^{\frac{5}{2}}=C \\ \Rightarrow x^8{\left|\frac{x^2-4y^2}{x^2}\right|}^{\frac{5}{2}}=C \\ \Rightarrow {\left|x^2-4y^2\right|}^5=C{x}^2 \\ (y(1)=0\Rightarrow |1{|}^5=C\Rightarrow C=1) \\ \therefore {\left|x^2-4y^2\right|}^5=x^2 \end{gathered}$$

12. Let y = y(x) be the solution of the differential equation $\frac{dy}{dx}=2x(x+y{)}^3-x(x+y)-1,y(0)=1$ . Then, ${\left(\frac{1}{\sqrt{2}}+\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2}\right)}^2$equals :

$$\begin{gathered} (1)\frac{4}{4+\sqrt{e}} \\ (2)\frac{3}{3-\sqrt{e}} \\ (3)\frac{2}{1+\sqrt{e}} \\ (4)\frac{1}{2-\sqrt{e}} \end{gathered}$$

Ans. (4)

Sol.

$$\begin{gathered} \frac{dy}{dx}=2x(x+y{)}^3-x(x+y)-1 \\ (x+y=t) \\ \Rightarrow \frac{dt}{dx}=1+\frac{dy}{dx} \\ \Rightarrow \frac{dt}{dx}=2x{t}^3-xt-1+1=2x{t}^3-xt \\ \Rightarrow \frac{dt}{2t^3-t}=xdx \\ \frac{t,dt}{2t^4-t^2}=x,dx \\ (t^2=z)， \\ \int \frac{dz}{2(2z^2-z)}=\int xdx \\ \int \frac{dz}{4z(z-\frac{1}{2})}=\int xdx \\ \ln \left|\frac{2z-1}{z}\right|=x^2+k \\ z=\frac{1}{2-\sqrt{e}} \end{gathered}$$

13. If the solution curve, of the differential equation $\frac{dy}{dx}=\frac{x+y-2}{x-y}$ passing through the point (2, 1) is ${\tan }^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta }{\log }_e\left(\alpha +{\left(\frac{y-1}{x-1}\right)}^2\right)={\log }_e|x-1|$ then $5\beta +\alpha$

is equal to

Ans. (11)

Sol.

$$\begin{gathered} \frac{dy}{dx}=\frac{x+y-2}{x-y} \\ x=X+h,y=Y+k \\ \frac{dY}{dX}=\frac{X+Y}{X-Y} \\ h+k-2=0,h-k=0\Rightarrow h=k=1 \\ (Y=vX) \\ v+\frac{dv}{dX}\cdot \frac{1+v}{1-v}=X-\frac{dv}{dX}\cdot \frac{1+v^2}{1-v} \\ \frac{1-v}{1+v^2}\cdot dv=\frac{dX}{X} \\ {\tan }^{-1}v-\frac{1}{2}\ln (1+v^2)=\ln |X|+C \\ \text{As curve is passing through }(2,1) \\ {\tan }^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2}\ln \left(1+{\left(\frac{y-1}{x-1}\right)}^2\right)=\ln |x-1| \\ \therefore \alpha =1\text{ and }\beta =2 \\ \Rightarrow 5\beta +\alpha =11 \end{gathered}$$

14. If $\sin \left(\frac{y}{x}\right)={\log }_e|x|+\frac{\alpha }{2}$ is the solution of the differential equation $x\cos \left(\frac{y}{x}\right)\frac{dy}{dx}=\cos \left(\frac{y}{x}\right)+x$ and y(1) = then $y(1)=\frac{\pi }{3},\text{ then }{\alpha }^2$ is equal to 

(1) 3

(2) 12

(3) 4

(4) 9

Ans. (1)

Sol.

Differential equation :–

$$\begin{gathered} x\cos \left(\frac{y}{x}\right)\frac{dy}{dx}=\cos \left(\frac{y}{x}\right)+x \\ \cos \left(\frac{y}{x}\right)\left[x\frac{dy}{dx}-y\right]=x \\ \text{Divide both sides by }{x}^2 \\ \cos \left(\frac{y}{x}\right)\left(\frac{x\frac{dy}{dx}-y}{x^2}\right)=\frac{1}{x} \\ \frac{y}{x}=t \\ \cos t\left(\frac{dt}{dx}\right)=\frac{1}{x} \\ \cos tdt=\frac{1}{x}dx \\ \sin t=\ln |x|+c \\ \sin \left(\frac{y}{x}\right)=\ln |x|+c \\ \text{Using }y(1)=\frac{\pi }{3},\text{ we get }c=\frac{\sqrt{3}}{2} \\ \text{So, }\alpha =\sqrt{3}\Rightarrow {\alpha }^2=3 \end{gathered}$$

15. Let Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y – y = Y′(x) (X – x) and the co-ordinate axes, where (x, y) is any point on the curve, is always $\frac{-y^2}{2Y^{\prime }(x)}+1,Y^{\prime }(x)\ne 0\text{. If }Y(1)=1$ , then 12Y(2) equals ______ .

Ans. (20)

Sol.

$$\begin{gathered} A=\frac{1}{2}\left(\frac{-y}{Y^{\prime }(x)}+x\left(\frac{y-xY/x}{x}\right)\right)=\frac{-y^2}{2Y^{\prime }(x)}+1 \\ (-y+x{Y}^{\prime }(x))(y-x{Y}^{\prime }(x))=-y^2+2Y^{\prime }(x) \\ -y^2+xy{Y}^{\prime }(x)+xy{Y}^{\prime }(x)-x^2[Y^{\prime }(x){]}^2 \\ =-y^2+2Y^{\prime }(x) \\ 2xy-x^2{Y}^{\prime }(x)=2 \\ \frac{dy}{dx}=\frac{2xy-2}{x^2} \\ \frac{dy}{dx}=\frac{2}{x}y-\frac{2}{x^2} \\ \text{I.F.}=e^{-2\ln x}=\frac{1}{x^2} \\ y\cdot \frac{1}{x^2}=\frac{2}{3}{x}^{-3}+c \\ \text{Put }x=1,y=1 \\ 1=\frac{2}{3}+c\Rightarrow c=\frac{1}{3} \\ Y=\frac{2}{3}\cdot \frac{1}{X}+\frac{1}{3}{X}^2 \\ 12Y(2)=\frac{5}{3}\cdot 12=20 \end{gathered}$$

16. Let y = y(x) be the solution of the differential equation $\Rightarrow \frac{dy}{dx}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x\tan x)},x\in \left(0,\frac{\pi }{2}\right)$ satisfying the condition $y\left(\frac{\pi }{4}\right)=2$ Then, $y\left(\frac{\pi }{3}\right)$ is

$$\begin{gathered} (1)\sqrt{3}(2+{\log }_e\sqrt{3}) \\ (2)\frac{\sqrt{3}}{2}(2+{\log }_{e}3) \\ (3)\sqrt{3}(1+2{\log }_{e}3) \\ (4)\sqrt{3}(2+{\log }_{e}3) \end{gathered}$$

Ans. (1)

Sol.

$$\begin{gathered} \frac{dy}{dx}=\frac{\sin x+y\cos x}{\sin x\cdot \cos x\left(\frac{1}{\cos x}-\sin x\cdot \frac{\sin x}{\cos x}\right)} \\ =\frac{\sin x+y\cos x}{\sin x(1-{\sin }^{2}x)} \\ \frac{dy}{dx}={\sec }^{2}x+y\cdot 2(\csc 2x) \\ \frac{dy}{dx}-2\csc (2x)\cdot y={\sec }^{2}x \\ \frac{dy}{dx}+py=Q \\ \text{I.F.}=e^{\int p,dx}=e^{\int -2\csc (2x),dx} \\ (2x=t) \\ 2\frac{dx}{dt}=1\Rightarrow dx=\frac{dt}{2} \\ =e^{-\int \csc (t),dt} \\ =e^{-\ln |\tan \frac{t}{2}|}=\frac{1}{|\tan \frac{t}{2}|} \\ =\frac{1}{|\tan x|}(\text{since }t=2x) \\ y(\text{I.F.})=\int Q(\text{I.F.})dx+c \\ \Rightarrow y\cdot \frac{1}{|\tan x|}=\int {\sec }^{2}x\cdot \frac{1}{|\tan x|}dx+c \\ y\cdot \frac{1}{|\tan x|}=\int \frac{dt}{|t|}+c\text{for }\tan x=t \\ y\cdot \frac{1}{|\tan x|}=\ln |t|+c \\ y=|\tan x|(\ln |\tan x|+c) \\ (x=\frac{\pi }{4},;y=2) \\ 2=\ln 1+c\Rightarrow c=2 \\ y=|\tan x|(\ln |\tan x|+2) \\ y\left(\frac{\pi }{3}\right)=\sqrt{3}(\ln \sqrt{3}+2) \end{gathered}$$

17. Let y = y(x) be the solution of the differential equation 

$$\begin{gathered} {\sec }^{2}xdx+\left(e^{2y}{\tan }^{2}x+\tan x\right)dy=0 \\ 0<x<\frac{\pi }{2},y\left(\frac{\pi }{4}\right)=0.\text{If }y\left(\frac{\pi }{6}\right)=\alpha \end{gathered}$$

Then $e^{3\alpha }$ is equal to ______.

Ans. (9)

Sol.

$$\begin{gathered} {\sec }^{2}x\cdot \frac{dx}{dy}+e^{2y}{\tan }^{2}x+\tan x=0 \\ (\tan x=t\Rightarrow {\sec }^{2}x\cdot \frac{dx}{dy}=\frac{dt}{dy}) \\ \Rightarrow \frac{dt}{dy}+e^{2y}{t}^2+t=0 \\ \frac{dt}{dy}+t=-t^2{e}^{2y} \\ \frac{1}{t^2}\cdot \frac{dt}{dy}+\frac{1}{t}=-e^{2y} \\ (\frac{1}{t}=u\Rightarrow \frac{dt}{dy}=-\frac{1}{u^2}\cdot \frac{du}{dy}) \\ -\frac{du}{dy}+u=-e^{2y} \\ \frac{du}{dy}=u{e}^{2y} \\ \text{Integrating Factor: }I.F.=(e^{\int -1,dy}=e^{-y}) \\ u{e}^{-y}=\int e^{-y}\cdot e^{2y},dy \\ u{e}^{-y}=\int e^y,dy=e^y+c \\ \frac{1}{\tan x}=e^y+c \\ x=\frac{\pi }{4},y=0\Rightarrow \frac{1}{\tan \frac{\pi }{4}}=1=e^0+c\Rightarrow c=0 \\ x=\frac{\pi }{6},y=\alpha \\ \frac{1}{\tan \frac{\pi }{6}}=\sqrt{3}=e^{\alpha } \\ \Rightarrow e^{3\alpha }={\left(e^{\alpha }\right)}^3=(\sqrt{3}{)}^3=3\sqrt{3} \\ \Rightarrow e^{3\alpha }=9 \end{gathered}$$