What is the solution for y^{''}=4y^'-3y,y(0)=5,y^'(0)=10 ?

The solution for y^{''}=4y^'-3y,y(0)=5,y^'(0)=10 is y=(5-5/2)e^{3t}+5/2 e^t

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y^{''}=4y^'-3y,y(0)=5,y^'(0)=10

y^{^{''}} = 4 y^{^{'}} - 3 y, y (0) = 5, y^{^{'}} (0) = 10

y = (5 - \frac{5}{2}) e^{3 t} + \frac{5}{2} e^{t}

Solution steps

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

Solve linear ODE:

y = (5 - \frac{5}{2}) e^{3 t} + \frac{5}{2} e^{t}

y = (5 - \frac{5}{2}) e^{3 t} + \frac{5}{2} e^{t}

y^{^{''}} + 36 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} - 8 y^{^{'}} - 16 y = 0

16 y^{^{''}} - 72 y^{^{'}} + 81 y = 0, y (0) = 8

y^{^{'''}} - 2 y^{^{''}} - 40 y^{^{'}} - 64 y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 29 y = 0

What is the solution for y^{''}=4y^'-3y,y(0)=5,y^'(0)=10 ?
The solution for y^{''}=4y^'-3y,y(0)=5,y^'(0)=10 is y=(5-5/2)e^{3t}+5/2 e^t