What is the solution for y^{''}-4y^'=7x+1 ?

The solution for y^{''}-4y^'=7x+1 is y=c_{1}e^{4x}+c_{2}-(7x^2)/8-(11x)/(16)

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y^{''}-4y^'=7x+1

y^{^{''}} - 4 y^{^{'}} = 7 x + 1

y = c_{1} e^{4 x} + c_{2} - \frac{7 x ^{2}}{8} - \frac{11 x}{16}

Solution steps

y^{''} (x) - 4 y^{'} (x) = 7 x + 1

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{7 x ^{2}}{8} - \frac{11 x}{16}

y = c_{1} e^{4 x} + c_{2} - \frac{7 x ^{2}}{8} - \frac{11 x}{16}

\frac{\partial}{\partial x} (e^{4 x} cos (7 y))

\frac{d}{d y} (y^{- 1})

\frac{d}{d x} (x^{2} e^{x} - 3)

x \to 0 lim (x \cdot arctan (\frac{1}{x}))

\int (2 x^{3} + 3 x + 1)^{4} (6 x^{2} + 3) d x

What is the solution for y^{''}-4y^'=7x+1 ?
The solution for y^{''}-4y^'=7x+1 is y=c_{1}e^{4x}+c_{2}-(7x^2)/8-(11x)/(16)