What is the solution for y^'=4y+x^2+9 ?

The solution for y^'=4y+x^2+9 is y=-(x^2)/4-x/8-73/32+c_{1}e^{4x}

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y^'=4y+x^2+9

y^{^{'}} = 4 y + x^{2} + 9

y = - \frac{x ^{2}}{4} - \frac{x}{8} - \frac{73}{32} + c_{1} e^{4 x}

Solution steps

y^{'} (x) = 4 y + x^{2} + 9

Solve linear ODE:

y = - \frac{x ^{2}}{4} - \frac{x}{8} - \frac{73}{32} + c_{1} e^{4 x}

y = - \frac{x ^{2}}{4} - \frac{x}{8} - \frac{73}{32} + c_{1} e^{4 x}

\int (3^{x} - sec^{2} (x)) d x

\frac{d y}{d x} = - 4 x + 6 x^{2}

x \to 4 lim (x^{2} + 1)

e^{x}

f (x) = \frac{x ^{3}}{6} + \frac{1}{2 x}

What is the solution for y^'=4y+x^2+9 ?
The solution for y^'=4y+x^2+9 is y=-(x^2)/4-x/8-73/32+c_{1}e^{4x}