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Differential Equations question

2023 · 24 Jan · Shift 1 · Q29
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  5. /2023 · 24 Jan · Shift 1 · Q29

Differential Equations question

2023 · 24 Jan · Shift 1 · Q29

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation x3dy+(xy−1)dx=0,x>0,y(12)=3−e{x^3}dy + (xy - 1)dx = 0,x \gt 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}x3dy+(xy−1)dx=0,x>0,y(21​)=3−e. Then y (1) is equal to
  1. A
    2 −-− e
  2. B
    3
  3. C
    1
  4. D
    e
View written solutionFree

Correct answer: C

  1. Rewrite the differential equation

Given x3 dy+(xy−1) dx=0.x^3\,dy+(xy-1)\,dx=0.x3dy+(xy−1)dx=0.

Divide by dxdxdx: x3dydx+xy−1=0.x^3\frac{dy}{dx}+xy-1=0.x3dxdy​+xy−1=0. So, x3dydx=1−xyx^3\frac{dy}{dx}=1-xyx3dxdy​=1−xy dydx+1x2y=1x3,x>0.\frac{dy}{dx}+\frac{1}{x^2}y=\frac{1}{x^3}, \qquad x>0.dxdy​+x21​y=x31​,x>0.

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), with P(x)=1x2,Q(x)=1x3.P(x)=\frac{1}{x^2}, \qquad Q(x)=\frac{1}{x^3}.P(x)=x21​,Q(x)=x31​.

  1. Find the integrating factor

The integrating factor is I.F.=e∫P(x) dx=e∫x−2dx=e−1/x.\mathrm{I.F.}=e^{\int P(x)\,dx}=e^{\int x^{-2}dx}=e^{-1/x}.I.F.=e∫P(x)dx=e∫x−2dx=e−1/x.

  1. Multiply the equation by the integrating factor

Multiplying throughout by e−1/xe^{-1/x}e−1/x: e−1/xdydx+1x2e−1/xy=1x3e−1/x.e^{-1/x}\frac{dy}{dx}+\frac{1}{x^2}e^{-1/x}y=\frac{1}{x^3}e^{-1/x}.e−1/xdxdy​+x21​e−1/xy=x31​e−1/x.

The left side becomes ddx(ye−1/x)=1x3e−1/x.\frac{d}{dx}\left(ye^{-1/x}\right)=\frac{1}{x^3}e^{-1/x}.dxd​(ye−1/x)=x31​e−1/x.

So, ye−1/x=∫1x3e−1/x dx+C.ye^{-1/x}=\int \frac{1}{x^3}e^{-1/x}\,dx + C.ye−1/x=∫x31​e−1/xdx+C.

  1. Evaluate the integral

Let t=−1x.t=-\frac{1}{x}.t=−x1​. Then dt=1x2dx.dt=\frac{1}{x^2}dx.dt=x21​dx.

We rewrite the integral more directly by using u=1/xu=1/xu=1/x: u=1x  ⟹  du=−1x2dx.u=\frac{1}{x} \implies du=-\frac{1}{x^2}dx.u=x1​⟹du=−x21​dx. Also, 1x3dx=1x⋅1x2dx=u(−du).\frac{1}{x^3}dx=\frac{1}{x}\cdot \frac{1}{x^2}dx=u\left(-du\right).x31​dx=x1​⋅x21​dx=u(−du). Hence, ∫1x3e−1/xdx=−∫ue−udu.\int \frac{1}{x^3}e^{-1/x}dx=-\int ue^{-u}du.∫x31​e−1/xdx=−∫ue−udu.

Now, ∫ue−udu=−(u+1)e−u+C,\int ue^{-u}du=-(u+1)e^{-u}+C,∫ue−udu=−(u+1)e−u+C, therefore −∫ue−udu=(u+1)e−u+C.-\int ue^{-u}du=(u+1)e^{-u}+C.−∫ue−udu=(u+1)e−u+C. Substituting back u=1/xu=1/xu=1/x, ∫1x3e−1/xdx=(1x+1)e−1/x+C.\int \frac{1}{x^3}e^{-1/x}dx=\left(\frac{1}{x}+1\right)e^{-1/x}+C.∫x31​e−1/xdx=(x1​+1)e−1/x+C.

Thus, ye−1/x=(1x+1)e−1/x+C.ye^{-1/x}=\left(\frac{1}{x}+1\right)e^{-1/x}+C.ye−1/x=(x1​+1)e−1/x+C. Multiplying by e1/xe^{1/x}e1/x, y=1x+1+Ce1/x.y=\frac{1}{x}+1+Ce^{1/x}.y=x1​+1+Ce1/x.

  1. Use the initial condition

Given y(12)=3−e.y\left(\frac12\right)=3-e.y(21​)=3−e. Substitute x=12x=\frac12x=21​: 3−e=11/2+1+Ce1/(1/2)=2+1+Ce2.3-e=\frac{1}{1/2}+1+Ce^{1/(1/2)}=2+1+Ce^2.3−e=1/21​+1+Ce1/(1/2)=2+1+Ce2. So, 3−e=3+Ce23-e=3+Ce^23−e=3+Ce2 Ce2=−eCe^2=-eCe2=−e C=−e−1.C=-e^{-1}.C=−e−1.

Hence, y=1x+1−e1/x−1.y=\frac{1}{x}+1-e^{1/x-1}.y=x1​+1−e1/x−1.

  1. Find y(1)y(1)y(1)

y(1)=1+1−e1−1=2−1=1.y(1)=1+1-e^{1-1}=2-1=1.y(1)=1+1−e1−1=2−1=1.

  1. Check the options
  • A: 2−e2-e2−e ❌
  • B: 333 ❌
  • C: 111 ✅
  • D: eee ❌

Therefore, the correct answer is Option C.

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