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Differential Equations question

2025 · 7 Apr · Shift 1 · Q32
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  5. /2025 · 7 Apr · Shift 1 · Q32

Differential Equations question

2025 · 7 Apr · Shift 1 · Q32

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation x(x2+ex)dy+(ex(x−2)y−x3)dx=0,x>0x\left(x^2+e^x\right) d y+\left(\mathrm{e}^x(x-2) y-x^3\right) \mathrm{d} x=0, x\gt 0x(x2+ex)dy+(ex(x−2)y−x3)dx=0,x>0, passing through the point (1,0)(1,0)(1,0). Then y(2)y(2)y(2) is equal to :
  1. A
    22+e2\frac{2}{2+e^2}2+e22​
  2. B
    44−e2\frac{4}{4-e^2}4−e24​
  3. C
    44+e2\frac{4}{4+e^2}4+e24​
  4. D
    22−e2\frac{2}{2-e^2}2−e22​
View written solutionFree

Correct answer: C

  1. Rewrite the differential equation in standard linear form

Given

x(x2+ex) dy+(ex(x−2)y−x3) dx=0,x>0.x(x^2+e^x)\,dy + \big(e^x(x-2)y - x^3\big)\,dx = 0, \qquad x>0.x(x2+ex)dy+(ex(x−2)y−x3)dx=0,x>0.

Divide by dxdxdx:

x(x2+ex)dydx+ex(x−2)y−x3=0.x(x^2+e^x)\frac{dy}{dx} + e^x(x-2)y - x^3 = 0.x(x2+ex)dxdy​+ex(x−2)y−x3=0.

So,

x(x2+ex)dydx+ex(x−2)y=x3.x(x^2+e^x)\frac{dy}{dx} + e^x(x-2)y = x^3.x(x2+ex)dxdy​+ex(x−2)y=x3.

Now divide by x(x2+ex)x(x^2+e^x)x(x2+ex):

dydx+ex(x−2)x(x2+ex)y=x3x(x2+ex).\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2+e^x)}y = \frac{x^3}{x(x^2+e^x)}.dxdy​+x(x2+ex)ex(x−2)​y=x(x2+ex)x3​.

Hence,

dydx+ex(x−2)x(x2+ex)y=x2x2+ex.\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2+e^x)}y = \frac{x^2}{x^2+e^x}.dxdy​+x(x2+ex)ex(x−2)​y=x2+exx2​.

This is a linear differential equation:

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

with

P(x)=ex(x−2)x(x2+ex).P(x)=\frac{e^x(x-2)}{x(x^2+e^x)}.P(x)=x(x2+ex)ex(x−2)​.
  1. Find the integrating factor

We need

I.F.=e∫P(x) dx.\text{I.F.} = e^{\int P(x)\,dx}.I.F.=e∫P(x)dx.

Observe that

ddx(x2+ex)=2x+ex,\frac{d}{dx}(x^2+e^x)=2x+e^x,dxd​(x2+ex)=2x+ex,

and

ddx(x2+exx)=x(2x+ex)−(x2+ex)x2=x2+ex(x−1)x2,\frac{d}{dx}\left(\frac{x^2+e^x}{x}\right) =\frac{x(2x+e^x)-(x^2+e^x)}{x^2} =\frac{x^2+e^x(x-1)}{x^2},dxd​(xx2+ex​)=x2x(2x+ex)−(x2+ex)​=x2x2+ex(x−1)​,

which is not directly helpful. Instead, try

ddxln⁡(x2+exx2)=2x+exx2+ex−2x.\frac{d}{dx}\ln\left(\frac{x^2+e^x}{x^2}\right) =\frac{2x+e^x}{x^2+e^x}-\frac{2}{x}.dxd​ln(x2x2+ex​)=x2+ex2x+ex​−x2​.

Combine over a common denominator x(x2+ex)x(x^2+e^x)x(x2+ex):

x(2x+ex)−2(x2+ex)x(x2+ex)=2x2+xex−2x2−2exx(x2+ex)=ex(x−2)x(x2+ex).\frac{x(2x+e^x)-2(x^2+e^x)}{x(x^2+e^x)} =\frac{2x^2+xe^x-2x^2-2e^x}{x(x^2+e^x)} =\frac{e^x(x-2)}{x(x^2+e^x)}.x(x2+ex)x(2x+ex)−2(x2+ex)​=x(x2+ex)2x2+xex−2x2−2ex​=x(x2+ex)ex(x−2)​.

Thus,

P(x)=ddxln⁡(x2+exx2).P(x)=\frac{d}{dx}\ln\left(\frac{x^2+e^x}{x^2}\right).P(x)=dxd​ln(x2x2+ex​).

Therefore,

I.F.=x2+exx2.\text{I.F.}=\frac{x^2+e^x}{x^2}.I.F.=x2x2+ex​.
  1. Multiply the equation by the integrating factor

The equation becomes

x2+exx2dydx+x2+exx2⋅ex(x−2)x(x2+ex)y=x2+exx2⋅x2x2+ex.\frac{x^2+e^x}{x^2}\frac{dy}{dx} + \frac{x^2+e^x}{x^2}\cdot \frac{e^x(x-2)}{x(x^2+e^x)}y = \frac{x^2+e^x}{x^2}\cdot \frac{x^2}{x^2+e^x}.x2x2+ex​dxdy​+x2x2+ex​⋅x(x2+ex)ex(x−2)​y=x2x2+ex​⋅x2+exx2​.

So,

ddx(y⋅x2+exx2)=1.\frac{d}{dx}\left(y\cdot \frac{x^2+e^x}{x^2}\right)=1.dxd​(y⋅x2x2+ex​)=1.

Integrating,

y⋅x2+exx2=x+C.y\cdot \frac{x^2+e^x}{x^2}=x+C.y⋅x2x2+ex​=x+C.

Hence,

y=x2(x+C)x2+ex.y=\frac{x^2(x+C)}{x^2+e^x}.y=x2+exx2(x+C)​.
  1. Use the condition y(1)=0y(1)=0y(1)=0

Given that the curve passes through (1,0)(1,0)(1,0),

0=12(1+C)1+e.0=\frac{1^2(1+C)}{1+e}.0=1+e12(1+C)​.

Thus,

1+C=0  ⟹  C=−1.1+C=0 \implies C=-1.1+C=0⟹C=−1.

So the solution is

y=x2(x−1)x2+ex.y=\frac{x^2(x-1)}{x^2+e^x}.y=x2+exx2(x−1)​.
  1. Find y(2)y(2)y(2)

Substitute x=2x=2x=2:

y(2)=22(2−1)22+e2=44+e2.y(2)=\frac{2^2(2-1)}{2^2+e^2}=\frac{4}{4+e^2}.y(2)=22+e222(2−1)​=4+e24​.
  1. Compare with the options
y(2)=44+e2y(2)=\frac{4}{4+e^2}y(2)=4+e24​

which matches Option C.

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