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

The solution for y^'+5y=e^{(-2x)} is y= 1/(3e^{2x)}+(c_{1})/(e^{5x)}

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

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

y = \frac{1}{3 e ^{2 x}} + \frac{c _{1}}{e ^{5 x}}

Solution steps

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

Solve linear ODE:

y = \frac{1}{3 e ^{2 x}} + \frac{c _{1}}{e ^{5 x}}

y = \frac{1}{3 e ^{2 x}} + \frac{c _{1}}{e ^{5 x}}

p (θ) = 3 θ

y = (x + 1)^{x}

\int_{0}^{2} 2 π x (2 x^{2} - x^{3}) d x

\int \frac{1}{( a + b ) - ( a - b ) x ^{2}} d x

\int \frac{22}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

What is the solution for y^'+5y=e^{(-2x)} ?
The solution for y^'+5y=e^{(-2x)} is y= 1/(3e^{2x)}+(c_{1})/(e^{5x)}