What is the solution for 2y^{''}+5y^'-3y=0,y(0)=1,y^'(0)=25 ?

The solution for 2y^{''}+5y^'-3y=0,y(0)=1,y^'(0)=25 is y=8e^{t/2}-7e^{-3t}

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2y^{''}+5y^'-3y=0,y(0)=1,y^'(0)=25

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 25

y = 8 e^{\frac{t}{2}} - 7 e^{- 3 t}

Solution steps

2 y^{''} (t) + 5 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = 8 e^{\frac{t}{2}} - 7 e^{- 3 t}

y = 8 e^{\frac{t}{2}} - 7 e^{- 3 t}

\frac{\partial}{\partial x} (2 x y)

\int 4 x^{3} + 3 x^{2} d x

\int \frac{cos ( x )}{2 sin ( x )} d x

x \to 0 lim ((\frac{1}{5 x}) - (\frac{1}{e ^{5 x}} - 1))

3 (5^{x})

What is the solution for 2y^{''}+5y^'-3y=0,y(0)=1,y^'(0)=25 ?
The solution for 2y^{''}+5y^'-3y=0,y(0)=1,y^'(0)=25 is y=8e^{t/2}-7e^{-3t}