What is the solution for 2y^{''}-6y^'=5,y(0)=2,y^'(0)=1 ?

The solution for 2y^{''}-6y^'=5,y(0)=2,y^'(0)=1 is y=(2-25/18)e^{3t}+25/18-(5t)/6

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2y^{''}-6y^'=5,y(0)=2,y^'(0)=1

2 y^{^{''}} - 6 y^{^{'}} = 5, y (0) = 2, y^{^{'}} (0) = 1

y = (2 - \frac{25}{18}) e^{3 t} + \frac{25}{18} - \frac{5 t}{6}

Solution steps

2 y^{''} (t) - 6 y^{'} (t) = 5

Solve linear ODE:

y = (2 - \frac{25}{18}) e^{3 t} + \frac{25}{18} - \frac{5 t}{6}

y = (2 - \frac{25}{18}) e^{3 t} + \frac{25}{18} - \frac{5 t}{6}

y^{^{''}} + y = 1, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 5 y^{^{'}} + 6 y = 1, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 25 y = 17 sec^{2} (5 t)

y^{^{''}} + y^{^{'}} - 6 y = 3 x - 2

y^{^{''}} - 16 y^{^{'}} + 64 y = 2 e^{8 x}

What is the solution for 2y^{''}-6y^'=5,y(0)=2,y^'(0)=1 ?
The solution for 2y^{''}-6y^'=5,y(0)=2,y^'(0)=1 is y=(2-25/18)e^{3t}+25/18-(5t)/6