What is the solution for 4y^{''}+y^'-3y=0,y(0)=-7 ?

The solution for 4y^{''}+y^'-3y=0,y(0)=-7 is y=(-7-c_{2})e^{(3t)/4}+c_{2}e^{-t}

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4y^{''}+y^'-3y=0,y(0)=-7

4 y^{^{''}} + y^{^{'}} - 3 y = 0, y (0) = - 7

y = (- 7 - c_{2}) e^{\frac{3 t}{4}} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (- 7 - c_{2}) e^{\frac{3 t}{4}} + c_{2} e^{- t}

y = (- 7 - c_{2}) e^{\frac{3 t}{4}} + c_{2} e^{- t}

y^{^{''}} + 49 π^{2} y = 0

\frac{d ^{2} x}{d t ^{2}} + 15 \frac{d x}{d t} + 56 x = 0

4 f (x) = f^{^{''}} (x)

y^{^{''}} + 6 y^{^{'}} = 0, y (0) = - 4, y^{^{'}} (0) = 6

y^{^{''}} = 7 y^{^{'}} - 10 y, y (0) = 0, y^{^{'}} (0) = - 3

What is the solution for 4y^{''}+y^'-3y=0,y(0)=-7 ?
The solution for 4y^{''}+y^'-3y=0,y(0)=-7 is y=(-7-c_{2})e^{(3t)/4}+c_{2}e^{-t}