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Popular >

y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1

Solution

y^{^{''}} + 12 y^{^{'}} - \frac{25}{4} y = 0, y (0) = 1, y^{^{'}} (0) = 1

Solution

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

Solution steps

y^{''} (t) + 12 y^{'} (t) - \frac{25}{4} y = 0

Solve linear ODE:

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

Graph

Popular Examples

x^{''}+9x=0,x(0)=5,x^'(0)=-11

x^{^{''}} + 9 x = 0, x (0) = 5, x^{^{'}} (0) = - 11

y^{''''}+5y^{'''}-2y^{''}-3y^'+1/2 y=0

y^{^{''''}} + 5 y^{^{'''}} - 2 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

(2d^5-7d^4+12d^3+8d^2)y=0

(2 d^{5} - 7 d^{4} + 12 d^{3} + 8 d^{2}) y = 0

linear y^{''}+y^'-2y=0 linear

y^{^{''}} + y^{^{'}} - 2 y = 0

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1 is y=(1+1/26)e^{t/2}-1/26 e^{-(25t)/2}