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Popular >

y^{''}-y^'=e^tcos(t),y(0)=0,y^'(0)=0

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

y^{''}-10y^'+25y=e^{5x}ln(4+x^2)

y^{^{''}} - 10 y^{^{'}} + 25 y = e^{5 x} ln (4 + x^{2})

y^{''}-2y^'-3y=4e^x-9

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 9

y^{''}-10y^'+9y=5t,y(0)=-1,y^'(0)=2

y^{^{''}} - 10 y^{^{'}} + 9 y = 5 t, y (0) = - 1, y^{^{'}} (0) = 2

y^{''}-2y^'+2y=(e^x)/(sin(x))

y^{^{''}} - 2 y^{^{'}} + 2 y = \frac{e ^{x}}{sin ( x )}

y^{^{''}} - y^{^{'}} = e^{x}

Frequently Asked Questions (FAQ)

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