What is the solution for 2y^{''}+5y^'-3y=0,y(0)=1,y^'(0)=4 ?

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

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2y^{''}+5y^'-3y=0,y(0)=1,y^'(0)=4

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 4

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

Solution steps

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

Solve linear ODE:

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

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

n = 0 \sum \infty \frac{1}{3 ^{n + 1}}

\int tan^{3} (2 t) sec^{3} (2 t) d t

\frac{d ^{2}}{d x ^{2}} (cos^{2} (2 π x))

\int x^{3} d x

ln (x^{2} + 1)

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