What is the solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-10 ?

The solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-10 is y=e^{4t}(a+(-10-4a)/5)-(e^{-t}(-10-4a))/5

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y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-10

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 10

y = e^{4 t} (a + \frac{- 10 - 4 a}{5}) - \frac{e ^{- t} ( - 10 - 4 a )}{5}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = e^{4 t} (a + \frac{- 10 - 4 a}{5}) - \frac{e ^{- t} ( - 10 - 4 a )}{5}

y = e^{4 t} (a + \frac{- 10 - 4 a}{5}) - \frac{e ^{- t} ( - 10 - 4 a )}{5}

y^{^{'}} - 2 y^{^{'''}} + y^{^{''}} = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 1

\frac{d ^{2} x}{d t ^{2}} + 0.75 \frac{d x}{d t} + 0.3 x = 0

y^{^{''}} + 10 y^{^{'}} + 2 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - \frac{5}{3}

What is the solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-10 ?
The solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-10 is y=e^{4t}(a+(-10-4a)/5)-(e^{-t}(-10-4a))/5