Let y = y(x) be the solution of the differential equation x\tan ( y \over x )dy = ( y\tan ( y \over x ) - x )dx , - 1 \le x \le 1 , y( 1 \over 2 ) = \pi \over 6 . Then the area of the region bounded by the curves x = 0, x = 1 \over \sqrt 2 and y = y(x) in the upper half plane is :

Differential Equations PYQ — JEE Main 2021 · 20 Jul · Shift 1 · Q29 with Solution

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Correct answer: A

Rewrite the differential equation

Given $x\tan\left(\frac yx\right)dy=\left(y\tan\left(\frac yx\right)-x\right)dx.$

Divide by $dx$ : $x\tan\left(\frac yx\right)\frac{dy}{dx}=y\tan\left(\frac yx\right)-x.$

So, $\frac{dy}{dx}=\frac{y}{x}-\cot\left(\frac yx\right).$

This is a homogeneous differential equation.

Substitute $v=\dfrac yx$

Let $y=vx \quad \Rightarrow \quad \frac{dy}{dx}=v+x\frac{dv}{dx}.$

Substitute into the differential equation: $v+x\frac{dv}{dx}=v-\cot v.$

Hence, $x\frac{dv}{dx}=-\cot v.$

So, $\frac{dv}{dx}=-\frac{\cot v}{x}.$

Separate variables: $\tan v\,dv=-\frac{dx}{x}.$

Integrate: $\int \tan v\,dv=\int -\frac{dx}{x}.$

Using $\int \tan v\,dv=-\ln|\cos v|$ , $-\ln|\cos v|=-\ln|x|+C.$

Thus, $\ln|\cos v|=\ln|x|+C_1,$ so $\cos v=Cx.$

Since $v=\dfrac yx$ , $\cos\left(\frac yx\right)=Cx.$

Use the initial condition

Given $y\left(\frac12\right)=\frac\pi6.$

Then $v=\frac{y}{x}=\frac{\pi/6}{1/2}=\frac\pi3.$

So, $\cos\left(\frac\pi3\right)=C\cdot \frac12.$

Since $\cos(\pi/3)=1/2$ , $\frac12=\frac C2 \Rightarrow C=1.$

Hence the solution is $\cos\left(\frac yx\right)=x.$

For the upper half plane and the given point, we take $\frac yx=\cos^{-1}x,$ so $y=x\cos^{-1}x.$

Find the required area

The region is bounded by $x=0$ , $x=\frac1{\sqrt2}$ , and $y=y(x)$ in the upper half plane.

Therefore, $A=\int_0^{1/\sqrt2} y\,dx=\int_0^{1/\sqrt2} x\cos^{-1}x\,dx.$

Evaluate the integral

Let $I=\int x\cos^{-1}x\,dx.$

Use integration by parts:

$u=\cos^{-1}x \Rightarrow du=-\dfrac{dx}{\sqrt{1-x^2}}$
$dv=x\,dx \Rightarrow v=\dfrac{x^2}{2}$

Then $I=\frac{x^2}{2}\cos^{-1}x+\frac12\int \frac{x^2}{\sqrt{1-x^2}}dx.$

Now, $x^2=1-(1-x^2),$ so $\frac{x^2}{\sqrt{1-x^2}}=\frac1{\sqrt{1-x^2}}-\sqrt{1-x^2}.$

Thus,

=\left[\frac{x^2}{2}\cos^{-1}x\right]_0^{1/\sqrt2} +\frac12\int_0^{1/\sqrt2}\frac{dx}{\sqrt{1-x^2}} -\frac12\int_0^{1/\sqrt2}\sqrt{1-x^2}\,dx.$$ Now evaluate each term: - First term: $$\left[\frac{x^2}{2}\cos^{-1}x\right]_0^{1/\sqrt2} =\frac{1/2}{2}\cdot \frac\pi4=\frac\pi{16}.$$ - Second term: $$\frac12\int_0^{1/\sqrt2}\frac{dx}{\sqrt{1-x^2}}= rac12\left[\sin^{-1}x\right]_0^{1/\sqrt2}= rac12\cdot\frac\pi4=\frac\pi8.$$ - Third term: Using $$\int \sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x,$$ we get $$\int_0^{1/\sqrt2}\sqrt{1-x^2}\,dx =\left[\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x\right]_0^{1/\sqrt2} =\frac14+\frac\pi8.$$ Therefore, $$A=\frac\pi{16}+\frac\pi8-\frac12\left(\frac14+\frac\pi8\right).$$ Simplify: $$A=\frac\pi{16}+\frac\pi8-\frac18-\frac\pi{16} =\frac\pi8-\frac18 =\frac18(\pi-1).$$ --- 6. **Match with the options** $$\boxed{A=\frac18(\pi-1)}$$ So the correct option is **A**.

More from Differential Equations

View written solutionFree

Correct answer: A

Rewrite the differential equation

Given $x\tan\left(\frac yx\right)dy=\left(y\tan\left(\frac yx\right)-x\right)dx.$

Divide by $dx$ : $x\tan\left(\frac yx\right)\frac{dy}{dx}=y\tan\left(\frac yx\right)-x.$

So, $\frac{dy}{dx}=\frac{y}{x}-\cot\left(\frac yx\right).$

This is a homogeneous differential equation.

Substitute $v=\dfrac yx$

Let $y=vx \quad \Rightarrow \quad \frac{dy}{dx}=v+x\frac{dv}{dx}.$

Substitute into the differential equation: $v+x\frac{dv}{dx}=v-\cot v.$

Hence, $x\frac{dv}{dx}=-\cot v.$

So, $\frac{dv}{dx}=-\frac{\cot v}{x}.$

Separate variables: $\tan v\,dv=-\frac{dx}{x}.$

Integrate: $\int \tan v\,dv=\int -\frac{dx}{x}.$

Using $\int \tan v\,dv=-\ln|\cos v|$ , $-\ln|\cos v|=-\ln|x|+C.$

Thus, $\ln|\cos v|=\ln|x|+C_1,$ so $\cos v=Cx.$

Since $v=\dfrac yx$ , $\cos\left(\frac yx\right)=Cx.$

Use the initial condition

Given $y\left(\frac12\right)=\frac\pi6.$

Then $v=\frac{y}{x}=\frac{\pi/6}{1/2}=\frac\pi3.$

So, $\cos\left(\frac\pi3\right)=C\cdot \frac12.$

Since $\cos(\pi/3)=1/2$ , $\frac12=\frac C2 \Rightarrow C=1.$

Hence the solution is $\cos\left(\frac yx\right)=x.$

For the upper half plane and the given point, we take $\frac yx=\cos^{-1}x,$ so $y=x\cos^{-1}x.$

Find the required area

The region is bounded by $x=0$ , $x=\frac1{\sqrt2}$ , and $y=y(x)$ in the upper half plane.

Therefore, $A=\int_0^{1/\sqrt2} y\,dx=\int_0^{1/\sqrt2} x\cos^{-1}x\,dx.$

Evaluate the integral

Let $I=\int x\cos^{-1}x\,dx.$

Use integration by parts:

$u=\cos^{-1}x \Rightarrow du=-\dfrac{dx}{\sqrt{1-x^2}}$
$dv=x\,dx \Rightarrow v=\dfrac{x^2}{2}$

Then $I=\frac{x^2}{2}\cos^{-1}x+\frac12\int \frac{x^2}{\sqrt{1-x^2}}dx.$

Now, $x^2=1-(1-x^2),$ so $\frac{x^2}{\sqrt{1-x^2}}=\frac1{\sqrt{1-x^2}}-\sqrt{1-x^2}.$

Thus,

=\left[\frac{x^2}{2}\cos^{-1}x\right]_0^{1/\sqrt2} +\frac12\int_0^{1/\sqrt2}\frac{dx}{\sqrt{1-x^2}} -\frac12\int_0^{1/\sqrt2}\sqrt{1-x^2}\,dx.$$ Now evaluate each term: - First term: $$\left[\frac{x^2}{2}\cos^{-1}x\right]_0^{1/\sqrt2} =\frac{1/2}{2}\cdot \frac\pi4=\frac\pi{16}.$$ - Second term: $$\frac12\int_0^{1/\sqrt2}\frac{dx}{\sqrt{1-x^2}}= rac12\left[\sin^{-1}x\right]_0^{1/\sqrt2}= rac12\cdot\frac\pi4=\frac\pi8.$$ - Third term: Using $$\int \sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x,$$ we get $$\int_0^{1/\sqrt2}\sqrt{1-x^2}\,dx =\left[\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x\right]_0^{1/\sqrt2} =\frac14+\frac\pi8.$$ Therefore, $$A=\frac\pi{16}+\frac\pi8-\frac12\left(\frac14+\frac\pi8\right).$$ Simplify: $$A=\frac\pi{16}+\frac\pi8-\frac18-\frac\pi{16} =\frac\pi8-\frac18 =\frac18(\pi-1).$$ --- 6. **Match with the options** $$\boxed{A=\frac18(\pi-1)}$$ So the correct option is **A**.

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