What is the solution for y^{''}+3y^'-18y=0,y(0)=a,y^'(0)=-12 ?

The solution for y^{''}+3y^'-18y=0,y(0)=a,y^'(0)=-12 is y=e^{3t}(a-(a+4)/3)+(e^{-6t}(a+4))/3

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y^{''}+3y^'-18y=0,y(0)=a,y^'(0)=-12

y^{^{''}} + 3 y^{^{'}} - 18 y = 0, y (0) = a, y^{^{'}} (0) = - 12

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

Solution steps

y^{''} (t) + 3 y^{'} (t) - 18 y = 0

Solve linear ODE:

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

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

(4 D^{2} - 4 D + 1) y = 0

u^{^{''}} + 4 H (t) = 0

x^{^{''}} - (\frac{π}{2})^{2} x = 0

y^{^{''}} - 7 y + 10 y = 0

3 y^{^{''}} + 27 y^{^{'}} + 24 y = 0

What is the solution for y^{''}+3y^'-18y=0,y(0)=a,y^'(0)=-12 ?
The solution for y^{''}+3y^'-18y=0,y(0)=a,y^'(0)=-12 is y=e^{3t}(a-(a+4)/3)+(e^{-6t}(a+4))/3