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

The solution for y^{''}-3y^'-10y=4,y(0)=0,y^'(0)=0 is y=(-2/7+2/5)e^{5t}+2/7 e^{-2t}-2/5

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

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

y = (- \frac{2}{7} + \frac{2}{5}) e^{5 t} + \frac{2}{7} e^{- 2 t} - \frac{2}{5}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 10 y = 4

Solve linear ODE:

y = (- \frac{2}{7} + \frac{2}{5}) e^{5 t} + \frac{2}{7} e^{- 2 t} - \frac{2}{5}

y = (- \frac{2}{7} + \frac{2}{5}) e^{5 t} + \frac{2}{7} e^{- 2 t} - \frac{2}{5}

\int_{1}^{e^{7}} \frac{ln ^{2} ( x ^{2} )}{x} d x

n = 1 \sum \infty (\frac{3}{n})^{4}

\int \frac{1}{sin ^{2} ( t )} d t

r = 1 0^{2 t}

\frac{\partial}{\partial x} (sin (π (4 x - 3 y)))

What is the solution for y^{''}-3y^'-10y=4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-3y^'-10y=4,y(0)=0,y^'(0)=0 is y=(-2/7+2/5)e^{5t}+2/7 e^{-2t}-2/5