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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{1}{x ^{7} + 2})

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

\int 3 e^{x} sin (3 x) d x

\frac{\partial}{\partial x} (sin (x) + y)

\int 2 t 1 + t^{2} d t

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