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Popular >

y^{''}-2y^'+5y=4e^xsin(2x)+5x

Solution

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 e^{x} sin (2 x) + 5 x

Solution

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = 4 e^{x} sin (2 x) + 5 x

Solve linear ODE:

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

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

Graph

Popular Examples

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

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

y^{''}-2y^'+y=te^{2t}

y^{^{''}} - 2 y^{^{'}} + y = t e^{2 t}

y^{'''}+y^{''}+y^'+y=e^{-t}+5t

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 5 t

(d^2y)/(dx^2)(1)+(dy)/(dx)(10)+9y=18

\frac{d ^{2} y}{d x ^{2}} (1) + \frac{d y}{d x} (10) + 9 y = 18

y^{''}+2y^'+4y=tcos(t)

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

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-2y^'+5y=4e^xsin(2x)+5x ?
The solution for y^{''}-2y^'+5y=4e^xsin(2x)+5x is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))-e^xxcos(2x)+x+2/5