What is the solution for y^{''}-10y^'+25y=(x+1)e^x ?

The solution for y^{''}-10y^'+25y=(x+1)e^x is y=c_{1}e^{5x}+c_{2}xe^{5x}+(e^xx)/(16)+3/32 e^x

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y^{''}-10y^'+25y=(x+1)e^x

y^{^{''}} - 10 y^{^{'}} + 25 y = (x + 1) e^{x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{e ^{x} x}{16} + \frac{3}{32} e^{x}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = (x + 1) e^{x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{e ^{x} x}{16} + \frac{3}{32} e^{x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{e ^{x} x}{16} + \frac{3}{32} e^{x}

y^{^{''}} - 6 y^{^{'}} + 8 y = x

y^{^{'''}} - 5 y^{^{''}} = 4 x^{2} - 3 x + 6 e^{5 x}

y^{^{''}} - 2 y^{^{'}} + y = 3 e^{2 t}

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

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

What is the solution for y^{''}-10y^'+25y=(x+1)e^x ?
The solution for y^{''}-10y^'+25y=(x+1)e^x is y=c_{1}e^{5x}+c_{2}xe^{5x}+(e^xx)/(16)+3/32 e^x