What is the solution for y^{''}-8y^'+15y=te^{3t} ?

The solution for y^{''}-8y^'+15y=te^{3t} is y=c_{1}e^{5t}+c_{2}e^{3t}-(e^{3t}t^2)/4-(e^{3t}t)/4

Popular >

y^{''}-8y^'+15y=te^{3t}

y^{^{''}} - 8 y^{^{'}} + 15 y = t e^{3 t}

y = c_{1} e^{5 t} + c_{2} e^{3 t} - \frac{e ^{3 t} t ^{2}}{4} - \frac{e ^{3 t} t}{4}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 15 y = t e^{3 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{3 t} - \frac{e ^{3 t} t ^{2}}{4} - \frac{e ^{3 t} t}{4}

y = c_{1} e^{5 t} + c_{2} e^{3 t} - \frac{e ^{3 t} t ^{2}}{4} - \frac{e ^{3 t} t}{4}

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

y^{^{''}} + k^{2} y = a cos (b x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} (\frac{1}{sin ( x )})

y^{^{''}} - y^{^{'}} - 6 y = 5 t, y (0) = 3, y^{^{'}} (0) = 1

D, D (x) = x^{8} - 4^{4}

What is the solution for y^{''}-8y^'+15y=te^{3t} ?
The solution for y^{''}-8y^'+15y=te^{3t} is y=c_{1}e^{5t}+c_{2}e^{3t}-(e^{3t}t^2)/4-(e^{3t}t)/4