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y^{''}-2y^'=cos(t),y(0)=0,y^'(0)=1

Solution

y^{^{''}} - 2 y^{^{'}} = cos (t), y (0) = 0, y^{^{'}} (0) = 1

Solution

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

Solution steps

y^{''} (t) - 2 y^{'} (t) = cos (t)

Solve linear ODE:

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

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

Graph

Popular Examples

y^{''}-11y^'+18y=6

y^{^{''}} - 11 y^{^{'}} + 18 y = 6

(d^2y)/(dx^2)+(dy)/(dx)+y=x^2-2x

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + y = x^{2} - 2 x

solvefor a,5d(a)=6d(a)+9d^2 solve for

a, 5 d (a) = 6 d (a) + 9 d^{2}

y^{''}+4y^'+4=e^{-2x}ln(x)

y^{^{''}} + 4 y^{^{'}} + 4 = e^{- 2 x} ln (x)

3/16 y^{''}+y=4cos(7t)+6

\frac{3}{16} y^{^{''}} + y = 4 cos (7 t) + 6

Frequently Asked Questions (FAQ)

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