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Popular >

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

Solution

y^{^{''}} - 2 y^{^{'}} + y = e^{x} \frac{1}{x ^{2} + 1}

Solution

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} e^{x} ln (x^{2} + 1) + e^{x} x arctan (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = e^{x} \frac{1}{x ^{2} + 1}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} e^{x} ln (x^{2} + 1) + e^{x} x arctan (x)

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} e^{x} ln (x^{2} + 1) + e^{x} x arctan (x)

Graph

Popular Examples

y^{''}-6y^'+9y=e^{3x}ln(x^2+1)

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} ln (x^{2} + 1)

y^{''}-6y^'+9y=t^2e^3

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3}

y^{''}-2y^'+y=2x+x^2+sin(3x)

y^{^{''}} - 2 y^{^{'}} + y = 2 x + x^{2} + sin (3 x)

y^{''}+4y=cos(2t),y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = cos (2 t), y (0) = 0, y^{^{'}} (0) = 0

y^{''}+y^'+5y=4e^{-t}cos(2t)

y^{^{''}} + y^{^{'}} + 5 y = 4 e^{- t} cos (2 t)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-2y^'+y=e^x 1/(x^2+1) ?
The solution for y^{''}-2y^'+y=e^x 1/(x^2+1) is y=c_{1}e^x+c_{2}xe^x-1/2 e^xln(x^2+1)+e^xxarctan(x)