What is the solution for y^'+4y=x^2 ?

The solution for y^'+4y=x^2 is y=(x^2)/4-(4x-1)/(32)+(c_{1})/(e^{4x)}

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y^'+4y=x^2

y^{^{'}} + 4 y = x^{2}

y = \frac{x ^{2}}{4} - \frac{4 x - 1}{32} + \frac{c _{1}}{e ^{4 x}}

Solution steps

y^{'} (x) + 4 y = x^{2}

Solve linear ODE:

y = \frac{x ^{2}}{4} - \frac{4 x - 1}{32} + \frac{c _{1}}{e ^{4 x}}

y = \frac{x ^{2}}{4} - \frac{4 x - 1}{32} + \frac{c _{1}}{e ^{4 x}}

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What is the solution for y^'+4y=x^2 ?
The solution for y^'+4y=x^2 is y=(x^2)/4-(4x-1)/(32)+(c_{1})/(e^{4x)}