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Differential Equations question

2024 · 6 Apr · Shift 1 · Q40
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  5. /2024 · 6 Apr · Shift 1 · Q40

Differential Equations question

2024 · 6 Apr · Shift 1 · Q40

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (2xlog⁡ex)dydx+2y=3xlog⁡ex,x>0\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x\gt 0(2xloge​x)dxdy​+2y=x3​loge​x,x>0 and y(e−1)=0y\left(e^{-1}\right)=0y(e−1)=0. Then, y(e)y(e)y(e) is equal to
  1. A
    −3e-\frac{3}{\mathrm{e}}−e3​
  2. B
    −32e-\frac{3}{2 \mathrm{e}}−2e3​
  3. C
    −23e-\frac{2}{3 \mathrm{e}}−3e2​
  4. D
    −2e-\frac{2}{\mathrm{e}}−e2​
View written solutionFree

Correct answer: A

  1. Given differential equation
(2xln⁡x)dydx+2y=3ln⁡xx,x>0(2x\ln x)\frac{dy}{dx}+2y=\frac{3\ln x}{x}, \qquad x>0(2xlnx)dxdy​+2y=x3lnx​,x>0

with condition

y(e−1)=0.y\left(e^{-1}\right)=0.y(e−1)=0.

We need to find y(e)y(e)y(e).

  1. Convert to standard linear form

Divide the whole equation by 2xln⁡x2x\ln x2xlnx:

dydx+1xln⁡xy=32x2.\frac{dy}{dx}+\frac{1}{x\ln x}y=\frac{3}{2x^2}.dxdy​+xlnx1​y=2x23​.

So this is a linear differential equation of the form

dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x)dxdy​+P(x)y=Q(x)

where

P(x)=1xln⁡x,Q(x)=32x2.P(x)=\frac{1}{x\ln x}, \qquad Q(x)=\frac{3}{2x^2}.P(x)=xlnx1​,Q(x)=2x23​.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(x)dx=e∫1xln⁡xdx.\text{I.F.}=e^{\int P(x)dx}=e^{\int \frac{1}{x\ln x}dx}.I.F.=e∫P(x)dx=e∫xlnx1​dx.

Since

∫1xln⁡xdx=ln⁡∣ln⁡x∣,\int \frac{1}{x\ln x}dx=\ln|\ln x|,∫xlnx1​dx=ln∣lnx∣,

we get

I.F.=eln⁡∣ln⁡x∣=∣ln⁡x∣.\text{I.F.}=e^{\ln|\ln x|}=|\ln x|.I.F.=eln∣lnx∣=∣lnx∣.

For our interval from x=e−1x=e^{-1}x=e−1 to x=ex=ex=e, the point x=1x=1x=1 is singular, so we solve on the branch containing the initial point x=e−1<1x=e^{-1}<1x=e−1<1, where ln⁡x<0\ln x<0lnx<0. A convenient integrating factor there is simply

I.F.=ln⁡x\text{I.F.}=\ln xI.F.=lnx

(up to a nonzero constant factor/sign, this is acceptable).

Multiplying the equation by ln⁡x\ln xlnx:

ln⁡x dydx+1xy=3ln⁡x2x2.\ln x\,\frac{dy}{dx}+\frac{1}{x}y=\frac{3\ln x}{2x^2}.lnxdxdy​+x1​y=2x23lnx​.

The left side is

ddx(yln⁡x).\frac{d}{dx}(y\ln x).dxd​(ylnx).

Hence,

ddx(yln⁡x)=3ln⁡x2x2.\frac{d}{dx}(y\ln x)=\frac{3\ln x}{2x^2}.dxd​(ylnx)=2x23lnx​.
  1. Integrate

Integrate both sides:

yln⁡x=∫3ln⁡x2x2 dx+C.y\ln x=\int \frac{3\ln x}{2x^2}\,dx + C.ylnx=∫2x23lnx​dx+C.

Now compute

I=∫ln⁡xx2dx.I=\int \frac{\ln x}{x^2}dx.I=∫x2lnx​dx.

Use integration by parts:

Let

u=ln⁡x,dw=1x2dx.u=\ln x, \qquad dw=\frac{1}{x^2}dx.u=lnx,dw=x21​dx.

Then

du=1xdx,w=−1x.du=\frac{1}{x}dx, \qquad w=-\frac{1}{x}.du=x1​dx,w=−x1​.

So

I=uv−∫w du=−ln⁡xx−∫(−1x⋅1x)dx=−ln⁡xx+∫1x2dx=−ln⁡xx−1x.I=uv-\int w\,du = -\frac{\ln x}{x}-\int \left(-\frac{1}{x}\cdot \frac{1}{x}\right)dx = -\frac{\ln x}{x}+\int \frac{1}{x^2}dx = -\frac{\ln x}{x}-\frac{1}{x}.I=uv−∫wdu=−xlnx​−∫(−x1​⋅x1​)dx=−xlnx​+∫x21​dx=−xlnx​−x1​.

Thus,

∫3ln⁡x2x2dx=32(−ln⁡xx−1x).\int \frac{3\ln x}{2x^2}dx=\frac{3}{2}\left(-\frac{\ln x}{x}-\frac{1}{x}\right).∫2x23lnx​dx=23​(−xlnx​−x1​).

Therefore,

yln⁡x=−32(ln⁡x+1x)+C.y\ln x=-\frac{3}{2}\left(\frac{\ln x+1}{x}\right)+C.ylnx=−23​(xlnx+1​)+C.

So,

y=−32ln⁡x+1x+Cln⁡x.y=\frac{-\frac{3}{2}\frac{\ln x+1}{x}+C}{\ln x}.y=lnx−23​xlnx+1​+C​.
  1. Use the initial condition

Given

y(e−1)=0.y(e^{-1})=0.y(e−1)=0.

At x=e−1x=e^{-1}x=e−1,

ln⁡(e−1)=−1,ln⁡x+1x=−1+1e−1=0.\ln(e^{-1})=-1, \qquad \frac{\ln x+1}{x}=\frac{-1+1}{e^{-1}}=0.ln(e−1)=−1,xlnx+1​=e−1−1+1​=0.

So

y(e−1)ln⁡(e−1)=0+C.y(e^{-1})\ln(e^{-1})=0+C.y(e−1)ln(e−1)=0+C.

Since y(e−1)=0y(e^{-1})=0y(e−1)=0, left side is 000, hence

C=0.C=0.C=0.

Thus,

yln⁡x=−32⋅ln⁡x+1x.y\ln x=-\frac{3}{2}\cdot \frac{\ln x+1}{x}.ylnx=−23​⋅xlnx+1​.
  1. Find y(e)y(e)y(e)

At x=ex=ex=e,

ln⁡e=1.\ln e=1.lne=1.

So

y(e)⋅1=−32⋅1+1e=−32⋅2e=−3e.y(e)\cdot 1=-\frac{3}{2}\cdot \frac{1+1}{e}=-\frac{3}{2}\cdot \frac{2}{e}=-\frac{3}{e}.y(e)⋅1=−23​⋅e1+1​=−23​⋅e2​=−e3​.

Hence,

y(e)=−3e.\boxed{y(e)=-\frac{3}{e}}.y(e)=−e3​​.
  1. Check options
  • A: −3e-\dfrac{3}{e}−e3​ ✅
  • B: −32e-\dfrac{3}{2e}−2e3​ ❌
  • C: −23e-\dfrac{2}{3e}−3e2​ ❌
  • D: −2e-\dfrac{2}{e}−e2​ ❌

So the correct option is A.

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