What is the solution for y^{''}-36y=12,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-36y=12,y(0)=1,y^'(0)=0 is y=(-2/3+4/3)e^{6t}+2/3 e^{-6t}-1/3

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y^{''}-36y=12,y(0)=1,y^'(0)=0

y^{^{''}} - 36 y = 12, y (0) = 1, y^{^{'}} (0) = 0

y = (- \frac{2}{3} + \frac{4}{3}) e^{6 t} + \frac{2}{3} e^{- 6 t} - \frac{1}{3}

Solution steps

y^{''} (t) - 36 y = 12

Solve linear ODE:

y = (- \frac{2}{3} + \frac{4}{3}) e^{6 t} + \frac{2}{3} e^{- 6 t} - \frac{1}{3}

y = (- \frac{2}{3} + \frac{4}{3}) e^{6 t} + \frac{2}{3} e^{- 6 t} - \frac{1}{3}

y^{^{''}} + 4 y^{^{'}} - y = k, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 25 y = sinh (5 x)

(D^{2} + 1) y = sin^{2} (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = - x e^{4} x

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} + 20 x = 5 H (t)

What is the solution for y^{''}-36y=12,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-36y=12,y(0)=1,y^'(0)=0 is y=(-2/3+4/3)e^{6t}+2/3 e^{-6t}-1/3