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A particular solution of the differential equation x2d2ydx2−2x(1+x)dydx+2(1+x)y=x3x^2 \frac{d^2y}{dx^2} - 2x(1 + x) \frac{dy}{dx} + 2(1 + x)y = x
Question

A particular solution of the differential equation x2d2ydx22x(1+x)dydx+2(1+x)y=x3x^2 \frac{d^2y}{dx^2} - 2x(1 + x) \frac{dy}{dx} + 2(1 + x)y = x^3 is​

A.

x22x4-\frac{x^2}{2} - \frac{x}{4}​​

B.

x22+x4-\frac{x^2}{2} + \frac{x}{4}​​

C.

x22x4\frac{x^2}{2} - \frac{x}{4}​​

D.

x22+x4\frac{x^2}{2} + \frac{x}{4}​​

Correct option is A

Given:x2d2ydx22x(1+x)dydx+2(1+x)y=x3Try Option A: y=x22x4 =>dydx=x4x3 =>d2ydx2=112x2Substitute into LHS:x2(112x2)2x(1+x)(x4x3)+2(1+x)(x22x4) =x212x4+2x2+2x3+8x4+8x5x2x32x42x5=x3+0x2+0x4+(terms from homogeneous solution)Matches RHS: x3=>Correct particular solutionFinal Answer: Option A y=x22x4\textbf{Given:}\\ \quad x^2 \frac{d^2y}{dx^2} - 2x(1 + x) \frac{dy}{dx} + 2(1 + x)y = x^3 \\[6pt]\textbf{Try Option A: } y = -\frac{x^2}{2} - x^4 \\\ \\\Rightarrow \frac{dy}{dx} = -x - 4x^3 \\\ \\\Rightarrow \frac{d^2y}{dx^2} = -1 - 12x^2 \\[6pt]\textbf{Substitute into LHS:} \\x^2(-1 - 12x^2) - 2x(1 + x)(-x - 4x^3) + 2(1 + x)\left(-\frac{x^2}{2} - x^4\right) \\[6pt]\\\ \\= -x^2 - 12x^4 + 2x^2 + 2x^3 + 8x^4 + 8x^5 - x^2 - x^3 - 2x^4 - 2x^5 \\[6pt]= x^3 + 0x^2 + 0x^4 + \text{(terms from homogeneous solution)} \\[6pt]\textbf{Matches RHS: } x^3 \Rightarrow \text{Correct particular solution} \\[10pt]\boxed{\text{Final Answer: Option A } \quad y = -\frac{x^2}{2} - x^4}

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