If the curve y = y(x) is the solution of the differential equation 2(x^2 + x^5/4)dy - y(x + x^1/4)dx = 2x^9/4dx , x > 0 which passes through the point ( 1,1 - 4 \over 3\log _e2 ) , then the value of y(16) is equal to :

4( 31 \over 3 - 8 \over 3\log _e3 )

( 31 \over 3 - 8 \over 3\log _e3 )

( 31 \over 3 + 8 \over 3\log _e3 )

4( 31 \over 3 + 8 \over 3\log _e3 )

Differential Equations PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q26 with Solution

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Correct answer: A

Rewrite the differential equation

Given $2(x^2+x^{5/4})\,dy-y(x+x^{1/4})\,dx=2x^{9/4}\,dx, \qquad x>0.$

Divide throughout by $dx$ : $2(x^2+x^{5/4})\frac{dy}{dx}-y(x+x^{1/4})=2x^{9/4}.$

Now simplify the coefficients: $x^2+x^{5/4}=x^{5/4}(x^{3/4}+1),$ $x+x^{1/4}=x^{1/4}(x^{3/4}+1).$

So the equation becomes $2x^{5/4}(x^{3/4}+1)\frac{dy}{dx}-yx^{1/4}(x^{3/4}+1)=2x^{9/4}.$

Since $x>0$ , divide by $x^{1/4}(x^{3/4}+1)$ : $2x\frac{dy}{dx}-y=\frac{2x^2}{x^{3/4}+1}.$

Hence $\frac{dy}{dx}-\frac{1}{2x}y=\frac{x}{x^{3/4}+1}.$

Solve the linear differential equation

This is of the form $\frac{dy}{dx}+P(x)y=Q(x),$ with $P(x)=-\frac{1}{2x}, \qquad Q(x)=\frac{x}{x^{3/4}+1}.$

The integrating factor is $\text{I.F.}=e^{\int -\frac{1}{2x}\,dx}=e^{-\frac12\ln x}=x^{-1/2}.$

Multiply the equation by $x^{-1/2}$ : $x^{-1/2}\frac{dy}{dx}-\frac{1}{2}x^{-3/2}y=\frac{x^{1/2}}{x^{3/4}+1}.$

The left side is $\frac{d}{dx}\left(yx^{-1/2}\right).$

So, $\frac{d}{dx}\left(yx^{-1/2}\right)=\frac{x^{1/2}}{x^{3/4}+1}.$

Integrate: $yx^{-1/2}=\int \frac{x^{1/2}}{x^{3/4}+1}\,dx+C.$

Evaluate the integral

Let $t=x^{1/4}\implies x=t^4, \quad dx=4t^3dt, \quad x^{1/2}=t^2, \quad x^{3/4}=t^3.$

Then

=\int \frac{t^2}{t^3+1}(4t^3dt) =4\int \frac{t^5}{t^3+1}\,dt.$$ Now divide: $$\frac{t^5}{t^3+1}=t^2-\frac{t^2}{t^3+1}.$$ Thus $$4\int \frac{t^5}{t^3+1}dt=4\int t^2dt-4\int \frac{t^2}{t^3+1}dt.$$ Compute each part: $$4\int t^2dt=\frac{4t^3}{3},$$ and for the second integral, let $u=t^3+1$, then $du=3t^2dt$: $$4\int \frac{t^2}{t^3+1}dt=\frac{4}{3}\int \frac{du}{u}=\frac{4}{3}\ln(t^3+1).$$ Hence $$\int \frac{x^{1/2}}{x^{3/4}+1}\,dx=\frac{4t^3}{3}-\frac{4}{3}\ln(t^3+1)+C.$$ Since $t=x^{1/4}$ and $t^3=x^{3/4}$, $$yx^{-1/2}=\frac{4}{3}x^{3/4}-\frac{4}{3}\ln(x^{3/4}+1)+C.$$ Therefore $$y= x^{1/2}\left(\frac{4}{3}x^{3/4}-\frac{4}{3}\ln(x^{3/4}+1)+C\right).$$ --- 4. **Use the given point** The curve passes through $$\left(1,1-\frac{4}{3}\ln 2\right).$$ Substitute $x=1$: $$1^{1/2}=1, \qquad 1^{3/4}=1.$$ So $$1-\frac{4}{3}\ln 2=\frac{4}{3}-\frac{4}{3}\ln 2+C.$$ Thus $$C=1-\frac{4}{3}=-\frac13.$$ Hence $$y= x^{1/2}\left(\frac{4}{3}x^{3/4}-\frac{4}{3}\ln(x^{3/4}+1)-\frac13\right).$$ --- 5. **Find $y(16)$** For $x=16$, $$16^{1/2}=4, \qquad 16^{3/4}=(16^{1/4})^3=2^3=8.$$ Therefore $$y(16)=4\left(\frac{4}{3}\cdot 8-\frac{4}{3}\ln(8+1)-\frac13\right).$$ Simplify: $$y(16)=4\left(\frac{32}{3}-\frac{4}{3}\ln 9-\frac13\right) =4\left(\frac{31}{3}-\frac{4}{3}\ln 9\right).$$ Since $$\ln 9=2\ln 3,$$ we get $$y(16)=4\left(\frac{31}{3}-\frac{8}{3}\ln 3\right).$$ --- 6. **Match with options** This is exactly **Option A**: $$\boxed{4\left(\frac{31}{3}-\frac{8}{3}\ln 3\right)}.$$

More from Differential Equations

View written solutionFree

Correct answer: A

Rewrite the differential equation

Given $2(x^2+x^{5/4})\,dy-y(x+x^{1/4})\,dx=2x^{9/4}\,dx, \qquad x>0.$

Divide throughout by $dx$ : $2(x^2+x^{5/4})\frac{dy}{dx}-y(x+x^{1/4})=2x^{9/4}.$

Now simplify the coefficients: $x^2+x^{5/4}=x^{5/4}(x^{3/4}+1),$ $x+x^{1/4}=x^{1/4}(x^{3/4}+1).$

So the equation becomes $2x^{5/4}(x^{3/4}+1)\frac{dy}{dx}-yx^{1/4}(x^{3/4}+1)=2x^{9/4}.$

Since $x>0$ , divide by $x^{1/4}(x^{3/4}+1)$ : $2x\frac{dy}{dx}-y=\frac{2x^2}{x^{3/4}+1}.$

Hence $\frac{dy}{dx}-\frac{1}{2x}y=\frac{x}{x^{3/4}+1}.$

Solve the linear differential equation

This is of the form $\frac{dy}{dx}+P(x)y=Q(x),$ with $P(x)=-\frac{1}{2x}, \qquad Q(x)=\frac{x}{x^{3/4}+1}.$

The integrating factor is $\text{I.F.}=e^{\int -\frac{1}{2x}\,dx}=e^{-\frac12\ln x}=x^{-1/2}.$

Multiply the equation by $x^{-1/2}$ : $x^{-1/2}\frac{dy}{dx}-\frac{1}{2}x^{-3/2}y=\frac{x^{1/2}}{x^{3/4}+1}.$

The left side is $\frac{d}{dx}\left(yx^{-1/2}\right).$

So, $\frac{d}{dx}\left(yx^{-1/2}\right)=\frac{x^{1/2}}{x^{3/4}+1}.$

Integrate: $yx^{-1/2}=\int \frac{x^{1/2}}{x^{3/4}+1}\,dx+C.$

Evaluate the integral

Let $t=x^{1/4}\implies x=t^4, \quad dx=4t^3dt, \quad x^{1/2}=t^2, \quad x^{3/4}=t^3.$

Then

=\int \frac{t^2}{t^3+1}(4t^3dt) =4\int \frac{t^5}{t^3+1}\,dt.$$ Now divide: $$\frac{t^5}{t^3+1}=t^2-\frac{t^2}{t^3+1}.$$ Thus $$4\int \frac{t^5}{t^3+1}dt=4\int t^2dt-4\int \frac{t^2}{t^3+1}dt.$$ Compute each part: $$4\int t^2dt=\frac{4t^3}{3},$$ and for the second integral, let $u=t^3+1$, then $du=3t^2dt$: $$4\int \frac{t^2}{t^3+1}dt=\frac{4}{3}\int \frac{du}{u}=\frac{4}{3}\ln(t^3+1).$$ Hence $$\int \frac{x^{1/2}}{x^{3/4}+1}\,dx=\frac{4t^3}{3}-\frac{4}{3}\ln(t^3+1)+C.$$ Since $t=x^{1/4}$ and $t^3=x^{3/4}$, $$yx^{-1/2}=\frac{4}{3}x^{3/4}-\frac{4}{3}\ln(x^{3/4}+1)+C.$$ Therefore $$y= x^{1/2}\left(\frac{4}{3}x^{3/4}-\frac{4}{3}\ln(x^{3/4}+1)+C\right).$$ --- 4. **Use the given point** The curve passes through $$\left(1,1-\frac{4}{3}\ln 2\right).$$ Substitute $x=1$: $$1^{1/2}=1, \qquad 1^{3/4}=1.$$ So $$1-\frac{4}{3}\ln 2=\frac{4}{3}-\frac{4}{3}\ln 2+C.$$ Thus $$C=1-\frac{4}{3}=-\frac13.$$ Hence $$y= x^{1/2}\left(\frac{4}{3}x^{3/4}-\frac{4}{3}\ln(x^{3/4}+1)-\frac13\right).$$ --- 5. **Find $y(16)$** For $x=16$, $$16^{1/2}=4, \qquad 16^{3/4}=(16^{1/4})^3=2^3=8.$$ Therefore $$y(16)=4\left(\frac{4}{3}\cdot 8-\frac{4}{3}\ln(8+1)-\frac13\right).$$ Simplify: $$y(16)=4\left(\frac{32}{3}-\frac{4}{3}\ln 9-\frac13\right) =4\left(\frac{31}{3}-\frac{4}{3}\ln 9\right).$$ Since $$\ln 9=2\ln 3,$$ we get $$y(16)=4\left(\frac{31}{3}-\frac{8}{3}\ln 3\right).$$ --- 6. **Match with options** This is exactly **Option A**: $$\boxed{4\left(\frac{31}{3}-\frac{8}{3}\ln 3\right)}.$$

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