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Popular >

15y^{''}+26y^'+7y=0,y(0)=8,y^'(0)=1

Solution

15 y^{^{''}} + 26 y^{^{'}} + 7 y = 0, y (0) = 8, y^{^{'}} (0) = 1

Solution

y = (8 + \frac{55}{16}) e^{- \frac{t}{3}} - \frac{55}{16} e^{- \frac{7 t}{5}}

Solution steps

15 y^{''} (t) + 26 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = (8 + \frac{55}{16}) e^{- \frac{t}{3}} - \frac{55}{16} e^{- \frac{7 t}{5}}

y = (8 + \frac{55}{16}) e^{- \frac{t}{3}} - \frac{55}{16} e^{- \frac{7 t}{5}}

Graph

Popular Examples

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

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

10y^{''''}+13y^{''}+4y=0

10 y^{^{''''}} + 13 y^{^{''}} + 4 y = 0

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

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

y^{''}+4y^'+4y=0,y(-1)=4,y^'(-1)=6

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 4, y^{^{'}} (- 1) = 6

(d^2y)/(dx^2)=ay

\frac{d ^{2} y}{d x ^{2}} = a y

Frequently Asked Questions (FAQ)

What is the solution for 15y^{''}+26y^'+7y=0,y(0)=8,y^'(0)=1 ?
The solution for 15y^{''}+26y^'+7y=0,y(0)=8,y^'(0)=1 is y=(8+55/16)e^{-t/3}-55/16 e^{-(7t)/5}