What is the solution for y^{''}-4y^'+5y=1+5x ?

The solution for y^{''}-4y^'+5y=1+5x is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+x+1

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y^{''}-4y^'+5y=1+5x

y^{^{''}} - 4 y^{^{'}} + 5 y = 1 + 5 x

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x + 1

Solution steps

y^{''} (x) - 4 y^{'} (x) + 5 y = 1 + 5 x

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x + 1

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x + 1

\int 9 x + 6 d x

\int \frac{x}{x ^{4} + 3 x ^{2} - 4} d x

n = 1 \sum \infty \frac{n + 2}{n !}

\frac{d N}{d t} = N (1 - 0.0004 N)

\int_{0}^{1} \frac{x}{x ^{2} + x + 1} d x

What is the solution for y^{''}-4y^'+5y=1+5x ?
The solution for y^{''}-4y^'+5y=1+5x is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+x+1