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

The solution for 2y^{''}-y^'-4y=0,y(2)=5,y^'(2)=0 is

Popular >

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

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

y = (5 - \frac{5 + 5 33}{2 33}) e^{\frac{( 1 + 33 ) t - 2 ( 1 + 33 )}{4}} + \frac{( 5 + 5 33 ) e ^{\frac{( 1 - 33 ) t - 2 ( 1 - 33 )}{4}}}{2 33}

Solution steps

2 y^{''} (t) - y^{'} (t) - 4 y = 0

Solve linear ODE:

y = (5 - \frac{5 + 5 33}{2 33}) e^{\frac{( 1 + 33 ) t - 2 ( 1 + 33 )}{4}} + \frac{( 5 + 5 33 ) e ^{\frac{( 1 - 33 ) t - 2 ( 1 - 33 )}{4}}}{2 33}

y = (5 - \frac{5 + 5 33}{2 33}) e^{\frac{( 1 + 33 ) t - 2 ( 1 + 33 )}{4}} + \frac{( 5 + 5 33 ) e ^{\frac{( 1 - 33 ) t - 2 ( 1 - 33 )}{4}}}{2 33}

y^{^{''}} - 6 y + 10 y = 0

y^{^{''}} - 7 y^{^{'}} = - 12 y

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

(D^{3} + 2 D^{2}) y = 0

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 7

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