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Differential Equations question

2024 · 4 Apr · Shift 1 · Q43
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  5. /2024 · 4 Apr · Shift 1 · Q43

Differential Equations question

2024 · 4 Apr · Shift 1 · Q43

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If the solution y=y(x)y=y(x)y=y(x) of the differential equation (x4+2x3+3x2+2x+2)dy−(2x2+2x+3)dx=0(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0(x4+2x3+3x2+2x+2)dy−(2x2+2x+3)dx=0 satisfies y(−1)=−π4y(-1)=-\frac{\pi}{4}y(−1)=−4π​, then y(0)y(0)y(0) is equal to :
  1. A
    −π12-\frac{\pi}{12}−12π​
  2. B
    π2\frac{\pi}{2}2π​
  3. C
    0
  4. D
    π4\frac{\pi}{4}4π​
View written solutionFree

Correct answer: D

  1. Rewrite the differential equation

Given

(x4+2x3+3x2+2x+2) dy−(2x2+2x+3) dx=0(x^4+2x^3+3x^2+2x+2)\,dy-(2x^2+2x+3)\,dx=0(x4+2x3+3x2+2x+2)dy−(2x2+2x+3)dx=0

so

(x4+2x3+3x2+2x+2)dydx=2x2+2x+3.(x^4+2x^3+3x^2+2x+2)\frac{dy}{dx}=2x^2+2x+3.(x4+2x3+3x2+2x+2)dxdy​=2x2+2x+3.

Hence,

dydx=2x2+2x+3x4+2x3+3x2+2x+2.\frac{dy}{dx}=\frac{2x^2+2x+3}{x^4+2x^3+3x^2+2x+2}.dxdy​=x4+2x3+3x2+2x+22x2+2x+3​.
  1. Factor the denominator

Observe that

x4+2x3+3x2+2x+2=(x2+1)(x2+2x+2).x^4+2x^3+3x^2+2x+2=(x^2+1)(x^2+2x+2).x4+2x3+3x2+2x+2=(x2+1)(x2+2x+2).

Indeed,

(x2+1)(x2+2x+2)=x4+2x3+3x2+2x+2.(x^2+1)(x^2+2x+2)=x^4+2x^3+3x^2+2x+2.(x2+1)(x2+2x+2)=x4+2x3+3x2+2x+2.

So,

dydx=2x2+2x+3(x2+1)(x2+2x+2).\frac{dy}{dx}=\frac{2x^2+2x+3}{(x^2+1)(x^2+2x+2)}.dxdy​=(x2+1)(x2+2x+2)2x2+2x+3​.
  1. Use partial fractions

Let

2x2+2x+3(x2+1)(x2+2x+2)=Ax+Bx2+1+Cx+Dx2+2x+2.\frac{2x^2+2x+3}{(x^2+1)(x^2+2x+2)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2x+2}.(x2+1)(x2+2x+2)2x2+2x+3​=x2+1Ax+B​+x2+2x+2Cx+D​.

Then

2x2+2x+3=(Ax+B)(x2+2x+2)+(Cx+D)(x2+1).2x^2+2x+3=(Ax+B)(x^2+2x+2)+(Cx+D)(x^2+1).2x2+2x+3=(Ax+B)(x2+2x+2)+(Cx+D)(x2+1).

Expanding,

(A+C)x3+(2A+B+D)x2+(2A+2B+C)x+(2B+D).(A+C)x^3+(2A+B+D)x^2+(2A+2B+C)x+(2B+D).(A+C)x3+(2A+B+D)x2+(2A+2B+C)x+(2B+D).

Comparing coefficients with 2x2+2x+32x^2+2x+32x2+2x+3, we get

A+C=0,A+C=0,A+C=0, 2A+B+D=2,2A+B+D=2,2A+B+D=2, 2A+2B+C=2,2A+2B+C=2,2A+2B+C=2, 2B+D=3.2B+D=3.2B+D=3.

From A+C=0A+C=0A+C=0, C=−AC=-AC=−A. Then

A+2B=2,A+2B=2,A+2B=2, 2A+B+D=2,2A+B+D=2,2A+B+D=2, 2B+D=3.2B+D=3.2B+D=3.

Subtracting the last two equations,

2A−B=−1.2A-B=-1.2A−B=−1.

Now solve

A+2B=2,qquad2A−B=−1.A+2B=2, qquad 2A-B=-1.A+2B=2,qquad2A−B=−1.

This gives

A=0,quadB=1.A=0, quad B=1.A=0,quadB=1.

Hence

C=0,quadD=1.C=0, quad D=1.C=0,quadD=1.

Therefore,

2x2+2x+3(x2+1)(x2+2x+2)=1x2+1+1x2+2x+2.\frac{2x^2+2x+3}{(x^2+1)(x^2+2x+2)}=\frac{1}{x^2+1}+\frac{1}{x^2+2x+2}.(x2+1)(x2+2x+2)2x2+2x+3​=x2+11​+x2+2x+21​.
  1. Integrate

So

dydx=1x2+1+1x2+2x+2.\frac{dy}{dx}=\frac{1}{x^2+1}+\frac{1}{x^2+2x+2}.dxdy​=x2+11​+x2+2x+21​.

Since

x2+2x+2=(x+1)2+1,x^2+2x+2=(x+1)^2+1,x2+2x+2=(x+1)2+1,

we get

y=∫dxx2+1+∫dx(x+1)2+1+C.y=\int \frac{dx}{x^2+1}+\int \frac{dx}{(x+1)^2+1}+C.y=∫x2+1dx​+∫(x+1)2+1dx​+C.

Thus,

y=tan⁡−1x+tan⁡−1(x+1)+C.y=\tan^{-1}x+\tan^{-1}(x+1)+C.y=tan−1x+tan−1(x+1)+C.
  1. Use the initial condition

Given

y(−1)=−π4.y(-1)=-\frac{\pi}{4}.y(−1)=−4π​.

Substitute x=−1x=-1x=−1:

y(−1)=tan⁡−1(−1)+tan⁡−1(0)+C=−π4+0+C.y(-1)=\tan^{-1}(-1)+\tan^{-1}(0)+C=-\frac{\pi}{4}+0+C.y(−1)=tan−1(−1)+tan−1(0)+C=−4π​+0+C.

So,

−π4+C=−π4  ⟹  C=0.-\frac{\pi}{4}+C=-\frac{\pi}{4} \implies C=0.−4π​+C=−4π​⟹C=0.

Hence,

y=tan⁡−1x+tan⁡−1(x+1).y=\tan^{-1}x+\tan^{-1}(x+1).y=tan−1x+tan−1(x+1).
  1. Find y(0)y(0)y(0)
y(0)=tan⁡−1(0)+tan⁡−1(1)=0+π4=π4.y(0)=\tan^{-1}(0)+\tan^{-1}(1)=0+\frac{\pi}{4}=\frac{\pi}{4}.y(0)=tan−1(0)+tan−1(1)=0+4π​=4π​.
  1. Compare with options

Thus,

y(0)=π4.y(0)=\frac{\pi}{4}.y(0)=4π​.

So the correct option is D.

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