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Popular >

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

Solution

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

Solution

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = 1 + x

Solve linear ODE:

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

Graph

Popular Examples

y^{''}-2y^'+y=x^3e^x

y^{^{''}} - 2 y^{^{'}} + y = x^{3} e^{x}

y^{^{''}} + 2 y - 1 = 0

y^{''}+5y^'+6y=6t+17,y(0)=3,y^'(0)=-4

y^{^{''}} + 5 y^{^{'}} + 6 y = 6 t + 17, y (0) = 3, y^{^{'}} (0) = - 4

y^{''}-3y=e^xsin(5x)

y^{^{''}} - 3 y = e^{x} sin (5 x)

2y^{''}-4y^'+10y=2,xy(0)=0,y^'(0)=0

2 y^{^{''}} - 4 y^{^{'}} + 10 y = 2, x y (0) = 0, y^{^{'}} (0) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-2y^'+5y=1+x,y(0)=0,y^'(0)=4 ?
The solution for y^{''}-2y^'+5y=1+x,y(0)=0,y^'(0)=4 is y=e^x(-7/25 cos(2x)+51/25 sin(2x))+x/5+7/25