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Differential Equations question

2024 · 1 Feb · Shift 1 · Q44
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  5. /2024 · 1 Feb · Shift 1 · Q44

Differential Equations question

2024 · 1 Feb · Shift 1 · Q44

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dy dx=2x(x+y)3−x(x+y)−1,y(0)=1\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1 dxdy​=2x(x+y)3−x(x+y)−1,y(0)=1. Then, (12+y(12))2\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2(2​1​+y(2​1​))2 equals :
  1. A
    44+e\frac{4}{4+\sqrt{\mathrm{e}}}4+e​4​
  2. B
    33−e\frac{3}{3-\sqrt{\mathrm{e}}}3−e​3​
  3. C
    21+e\frac{2}{1+\sqrt{\mathrm{e}}}1+e​2​
  4. D
    12−e\frac{1}{2-\sqrt{\mathrm{e}}}2−e​1​
View written solutionFree

Correct answer: D

  1. Given differential equation

We have

dydx=2x(x+y)3−x(x+y)−1,y(0)=1.\frac{dy}{dx}=2x(x+y)^3-x(x+y)-1, \qquad y(0)=1.dxdy​=2x(x+y)3−x(x+y)−1,y(0)=1.

We need to find

(12+y(12))2.\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2.(2​1​+y(2​1​))2.
  1. Use substitution

Let

z=x+y.z=x+y.z=x+y.

Then

dzdx=1+dydx.\frac{dz}{dx}=1+\frac{dy}{dx}.dxdz​=1+dxdy​.

Using the given DE,

dzdx=1+(2xz3−xz−1)=2xz3−xz.\frac{dz}{dx}=1+\left(2xz^3-xz-1\right)=2xz^3-xz.dxdz​=1+(2xz3−xz−1)=2xz3−xz.

So,

dzdx=xz(2z2−1).\frac{dz}{dx}=xz(2z^2-1).dxdz​=xz(2z2−1).

Also, from y(0)=1y(0)=1y(0)=1,

z(0)=0+y(0)=1.z(0)=0+y(0)=1.z(0)=0+y(0)=1.
  1. Transform to a simpler equation

We are asked for z2z^2z2 at x=12x=\frac{1}{\sqrt2}x=2​1​, so let

u=z2.u=z^2.u=z2.

Then

dνdx=2zdzdx=2z⋅xz(2z2−1)=2xν(2ν−1).\frac{d\nu}{dx}=2z\frac{dz}{dx}=2z\cdot xz(2z^2-1)=2x\nu(2\nu-1).dxdν​=2zdxdz​=2z⋅xz(2z2−1)=2xν(2ν−1).

Thus,

dνdx=2xν(2ν−1).\frac{d\nu}{dx}=2x\nu(2\nu-1).dxdν​=2xν(2ν−1).

Initial condition:

ν(0)=z(0)2=1.\nu(0)=z(0)^2=1.ν(0)=z(0)2=1.
  1. Separate variables
dνν(2ν−1)=2x dx.\frac{d\nu}{\nu(2\nu-1)}=2x\,dx.ν(2ν−1)dν​=2xdx.

Now do partial fractions:

1ν(2ν−1)=Aν+B2ν−1.\frac{1}{\nu(2\nu-1)}=\frac{A}{\nu}+\frac{B}{2\nu-1}.ν(2ν−1)1​=νA​+2ν−1B​.

So,

1=A(2ν−1)+Bν=(2A+B)ν−A.1=A(2\nu-1)+B\nu=(2A+B)\nu-A.1=A(2ν−1)+Bν=(2A+B)ν−A.

Comparing coefficients:

−A=1⇒A=−1,-A=1 \Rightarrow A=-1,−A=1⇒A=−1, 2A+B=0⇒−2+B=0⇒B=2.2A+B=0 \Rightarrow -2+B=0 \Rightarrow B=2.2A+B=0⇒−2+B=0⇒B=2.

Hence,

1ν(2ν−1)=−1ν+22ν−1.\frac{1}{\nu(2\nu-1)}=-\frac{1}{\nu}+\frac{2}{2\nu-1}.ν(2ν−1)1​=−ν1​+2ν−12​.

Therefore,

∫dνν(2ν−1)=∫(−1ν+22ν−1)dν=−ln⁡∣ν∣+ln⁡∣2ν−1∣.\int \frac{d\nu}{\nu(2\nu-1)}=\int\left(-\frac{1}{\nu}+\frac{2}{2\nu-1}\right)d\nu = -\ln|\nu|+\ln|2\nu-1|.∫ν(2ν−1)dν​=∫(−ν1​+2ν−12​)dν=−ln∣ν∣+ln∣2ν−1∣.

Thus,

ln⁡∣2ν−1ν∣=x2+C.\ln\left|\frac{2\nu-1}{\nu}\right|=x^2+C.ln​ν2ν−1​​=x2+C.
  1. Use the initial condition

At x=0x=0x=0, ν=1\nu=1ν=1, so

ln⁡(2(1)−11)=0+C.\ln\left(\frac{2(1)-1}{1}\right)=0+C.ln(12(1)−1​)=0+C.

That is,

ln⁡(1)=C=0.\ln(1)=C=0.ln(1)=C=0.

So,

ln⁡(2ν−1ν)=x2.\ln\left(\frac{2\nu-1}{\nu}\right)=x^2.ln(ν2ν−1​)=x2.

Exponentiating,

2ν−1ν=ex2.\frac{2\nu-1}{\nu}=e^{x^2}.ν2ν−1​=ex2.

Hence,

2ν−1=νex22\nu-1=\nu e^{x^2}2ν−1=νex2 ν(2−ex2)=1\nu(2-e^{x^2})=1ν(2−ex2)=1 ν=12−ex2.\nu=\frac{1}{2-e^{x^2}}.ν=2−ex21​.

Since ν=z2=(x+y)2\nu=z^2=(x+y)^2ν=z2=(x+y)2,

(x+y)2=12−ex2.(x+y)^2=\frac{1}{2-e^{x^2}}.(x+y)2=2−ex21​.
  1. Evaluate at x=12x=\frac{1}{\sqrt2}x=2​1​

Then

x2=12,x^2=\frac12,x2=21​,

so

(12+y(12))2=12−e1/2=12−e.\left(\frac{1}{\sqrt2}+y\left(\frac{1}{\sqrt2}\right)\right)^2 =\frac{1}{2-e^{1/2}} =\frac{1}{2-\sqrt e}.(2​1​+y(2​1​))2=2−e1/21​=2−e​1​.
  1. Compare with options

This matches

D: 12−e.\boxed{\text{D: } \frac{1}{2-\sqrt e}}.D: 2−e​1​​.
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