What is the solution for y^{''}-10y^'+25y=0,y(0)=xe^{5x} ?

The solution for y^{''}-10y^'+25y=0,y(0)=xe^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}

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y^{''}-10y^'+25y=0,y(0)=xe^{5x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = x e^{5 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 0

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 29 y = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + k y^{^{'}} + 8 y = 0, y (0) = e^{2 x} sin (2 x)

y^{^{''}} - 10 y = 0

What is the solution for y^{''}-10y^'+25y=0,y(0)=xe^{5x} ?
The solution for y^{''}-10y^'+25y=0,y(0)=xe^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}