What is the solution for y^{''}-8y^'+15y=0,y(0)=1,y^'(0)=4 ?

The solution for y^{''}-8y^'+15y=0,y(0)=1,y^'(0)=4 is y=(1-1/2)e^{5t}+1/2 e^{3t}

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y^{''}-8y^'+15y=0,y(0)=1,y^'(0)=4

y^{^{''}} - 8 y^{^{'}} + 15 y = 0, y (0) = 1, y^{^{'}} (0) = 4

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{3 t}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 15 y = 0

Solve linear ODE:

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{3 t}

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{3 t}

\frac{d y}{d t} + y = 2, y (0) = 0

\int cos^{4} (θ) sin (θ) d θ

\frac{d}{d x} (ln (3 x^{2} - 2 x))

\frac{d}{d x} (ln ((x^{2} + 1)^{3}))

\int (x + 3) 4 - x^{2} d x

What is the solution for y^{''}-8y^'+15y=0,y(0)=1,y^'(0)=4 ?
The solution for y^{''}-8y^'+15y=0,y(0)=1,y^'(0)=4 is y=(1-1/2)e^{5t}+1/2 e^{3t}