What is the solution for y^{''}-4y^'+5y=e^{2x} ?

The solution for y^{''}-4y^'+5y=e^{2x} is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+e^{2x}

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

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

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{2 x}

Solution steps

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

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{2 x}

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{2 x}

\frac{\partial}{\partial x} ((1 + x y) e^{x y})

h \to 0 lim (2 x^{2} + 3 x h + 2)

y^{^{'}} = \frac{( 3 e ^{x} )}{e ^{y}}

x tan (\frac{y}{x}) + y - x y^{^{'}} = 0

\int_{1}^{2} (x + \frac{1}{x ^{2}}) d x

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