What is the solution for y^{''}-8y^'+15y=2e^{3x} ?

The solution for y^{''}-8y^'+15y=2e^{3x} is y=c_{1}e^{5x}+c_{2}e^{3x}-e^{3x}x

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y^{''}-8y^'+15y=2e^{3x}

y^{^{''}} - 8 y^{^{'}} + 15 y = 2 e^{3 x}

y = c_{1} e^{5 x} + c_{2} e^{3 x} - e^{3 x} x

Solution steps

y^{''} (x) - 8 y^{'} (x) + 15 y = 2 e^{3 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{3 x} - e^{3 x} x

y = c_{1} e^{5 x} + c_{2} e^{3 x} - e^{3 x} x

e^{x}, x e^{x}, [0, 1]

\frac{\partial}{\partial x} (a ln (x) + b ln (y))

(t - 2)^{3}

- \frac{x}{2}

\int \frac{( 3 )}{( u ^{2} - 1 )} d u

What is the solution for y^{''}-8y^'+15y=2e^{3x} ?
The solution for y^{''}-8y^'+15y=2e^{3x} is y=c_{1}e^{5x}+c_{2}e^{3x}-e^{3x}x