What is the solution for y^{''}+4y^'-32y=0,y(0)=a,y^'(0)=-40 ?

The solution for y^{''}+4y^'-32y=0,y(0)=a,y^'(0)=-40 is y=e^{4t}(a-(a+10)/3)+(e^{-8t}(a+10))/3

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y^{''}+4y^'-32y=0,y(0)=a,y^'(0)=-40

y^{^{''}} + 4 y^{^{'}} - 32 y = 0, y (0) = a, y^{^{'}} (0) = - 40

y = e^{4 t} (a - \frac{a + 10}{3}) + \frac{e ^{- 8 t} ( a + 10 )}{3}

Solution steps

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

Solve linear ODE:

y = e^{4 t} (a - \frac{a + 10}{3}) + \frac{e ^{- 8 t} ( a + 10 )}{3}

y = e^{4 t} (a - \frac{a + 10}{3}) + \frac{e ^{- 8 t} ( a + 10 )}{3}

77 y^{^{''''}} + 303 y^{^{''}} + 100 y = 0

x^{^{''}} + 25 x = 0, x (0) = 0.6, x^{^{'}} (0) = 1

- a \cdot y^{^{''}} + y^{^{'}} = 0, y (0) = 1, y (1) = 0

16 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'-32y=0,y(0)=a,y^'(0)=-40 ?
The solution for y^{''}+4y^'-32y=0,y(0)=a,y^'(0)=-40 is y=e^{4t}(a-(a+10)/3)+(e^{-8t}(a+10))/3