What is the solution for (d^2y)/(dx^2)-5(dy)/(dx)+4y=e^x ?

The solution for (d^2y)/(dx^2)-5(dy)/(dx)+4y=e^x is y=c_{1}e^{4x}+c_{2}e^x-(e^xx)/3

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(d^2y)/(dx^2)-5(dy)/(dx)+4y=e^x

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = e^{x}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{e ^{x} x}{3}

Solution steps

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{e ^{x} x}{3}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{e ^{x} x}{3}

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 7 - e^{x} + e^{5 x}

(D^{2} + D - 2) y = e^{x}

y^{^{''}} - 8 y^{^{'}} + 15 y = e^{3 x} sin (3 x)

x^{^{''}} + a^{2} x = a^{2} b + c

y^{^{''}} - 10 y^{^{'}} + 9 y = 5

What is the solution for (d^2y)/(dx^2)-5(dy)/(dx)+4y=e^x ?
The solution for (d^2y)/(dx^2)-5(dy)/(dx)+4y=e^x is y=c_{1}e^{4x}+c_{2}e^x-(e^xx)/3