What is the solution for y^'=4y+x^2+7 ?

The solution for y^'=4y+x^2+7 is y=-(x^2)/4-x/8-57/32+c_{1}e^{4x}

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y^'=4y+x^2+7

y^{^{'}} = 4 y + x^{2} + 7

y = - \frac{x ^{2}}{4} - \frac{x}{8} - \frac{57}{32} + c_{1} e^{4 x}

Solution steps

y^{'} (x) = 4 y + x^{2} + 7

Solve linear ODE:

y = - \frac{x ^{2}}{4} - \frac{x}{8} - \frac{57}{32} + c_{1} e^{4 x}

y = - \frac{x ^{2}}{4} - \frac{x}{8} - \frac{57}{32} + c_{1} e^{4 x}

What is the solution for y^'=4y+x^2+7 ?
The solution for y^'=4y+x^2+7 is y=-(x^2)/4-x/8-57/32+c_{1}e^{4x}