What is the solution for y^{''}-6y^'+8y=e^{3t} ?

The solution for y^{''}-6y^'+8y=e^{3t} is y=c_{1}e^{4t}+c_{2}e^{2t}-e^{3t}

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y^{''}-6y^'+8y=e^{3t}

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 t}

y = c_{1} e^{4 t} + c_{2} e^{2 t} - e^{3 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = e^{3 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{2 t} - e^{3 t}

y = c_{1} e^{4 t} + c_{2} e^{2 t} - e^{3 t}

\int \frac{x ^{4} + x - 1}{x ^{2} + 1} d x

\int \frac{x ^{3} - 2 x + 3}{x ^{2} - x + 1} d x

\frac{\partial}{\partial x} ((1 - x) + 10 (y - x^{2})^{2})

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

\frac{d}{d x} (x - 2 - 0.5 sin (x))

What is the solution for y^{''}-6y^'+8y=e^{3t} ?
The solution for y^{''}-6y^'+8y=e^{3t} is y=c_{1}e^{4t}+c_{2}e^{2t}-e^{3t}