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

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

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

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

y = 2 e^{\frac{3 t - 10}{2} + 2} + 3 e^{2 - t}

Solution steps

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

Solve linear ODE:

y = 2 e^{\frac{3 t - 10}{2} + 2} + 3 e^{2 - t}

y = 2 e^{\frac{3 t - 10}{2} + 2} + 3 e^{2 - t}

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