What is the solution for y^{''}+4y^'-32y=0,y(0)=-3,y^'(0)=-12 ?

The solution for y^{''}+4y^'-32y=0,y(0)=-3,y^'(0)=-12 is y=-3e^{4t}

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y^{''}+4y^'-32y=0,y(0)=-3,y^'(0)=-12

y^{^{''}} + 4 y^{^{'}} - 32 y = 0, y (0) = - 3, y^{^{'}} (0) = - 12

y = - 3 e^{4 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 32 y = 0

Solve linear ODE:

y = - 3 e^{4 t}

y = - 3 e^{4 t}

10 y^{^{''}} + 6 y^{^{'}} + y = 0

u^{^{''}} + 14 H (t) = 0

9 y^{^{''}} + 30 y^{^{'}} - 25 y = 0

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} - 5 y = 0, y (1) = 0

x^{^{''}} = - a^{2} x

What is the solution for y^{''}+4y^'-32y=0,y(0)=-3,y^'(0)=-12 ?
The solution for y^{''}+4y^'-32y=0,y(0)=-3,y^'(0)=-12 is y=-3e^{4t}