What is the solution for 4y^{''}+3y^'=0,y(0)=-2,y^'(0)=3 ?

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

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4y^{''}+3y^'=0,y(0)=-2,y^'(0)=3

4 y^{^{''}} + 3 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = 3

y = - 4 e^{- \frac{3 t}{4}} + 2

Solution steps

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

Solve linear ODE:

y = - 4 e^{- \frac{3 t}{4}} + 2

y = - 4 e^{- \frac{3 t}{4}} + 2

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 1, y (1) = 0

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0

y^{^{'''}} - 7 y^{^{'}} - 6 y = 0

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

y^{^{'''}} - 4 y^{^{''}} + 4 y^{^{'}} = 0

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