What is the solution for y^{''}-49y=2xe^{7x} ?

The solution for y^{''}-49y=2xe^{7x} is y=c_{1}e^{7x}+c_{2}e^{-7x}+(e^{7x}x^2)/(14)-(e^{7x}x)/(98)

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y^{''}-49y=2xe^{7x}

y^{^{''}} - 49 y = 2 x e^{7 x}

y = c_{1} e^{7 x} + c_{2} e^{- 7 x} + \frac{e ^{7 x} x ^{2}}{14} - \frac{e ^{7 x} x}{98}

Solution steps

y^{''} (x) - 49 y = 2 x e^{7 x}

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} e^{- 7 x} + \frac{e ^{7 x} x ^{2}}{14} - \frac{e ^{7 x} x}{98}

y = c_{1} e^{7 x} + c_{2} e^{- 7 x} + \frac{e ^{7 x} x ^{2}}{14} - \frac{e ^{7 x} x}{98}

(D^{2} + 4) y = x sin (x)

y^{^{''}} + 4 y + 4 y = 3, e^{- t} y (0) = 2

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

d (x) = x + 8

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 sin (t) + 2 e^{2 t}

What is the solution for y^{''}-49y=2xe^{7x} ?
The solution for y^{''}-49y=2xe^{7x} is y=c_{1}e^{7x}+c_{2}e^{-7x}+(e^{7x}x^2)/(14)-(e^{7x}x)/(98)