What is the solution for y^'=-3y+e^{-2x},y(0)=0 ?

The solution for y^'=-3y+e^{-2x},y(0)=0 is y= 1/(e^{2x)}-1/(e^{3x)}

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y^'=-3y+e^{-2x},y(0)=0

y^{^{'}} = - 3 y + e^{- 2 x}, y (0) = 0

y = \frac{1}{e ^{2 x}} - \frac{1}{e ^{3 x}}

Solution steps

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

Solve linear ODE:

y = \frac{1}{e ^{2 x}} - \frac{1}{e ^{3 x}}

y = \frac{1}{e ^{2 x}} - \frac{1}{e ^{3 x}}

f (x) = \frac{2}{1 + x ^{2}}, h (x) = x^{2}

\frac{d y}{d x} = (\frac{( x y ) ^{2}}{8} + y^{2})

\frac{d ^{2}}{d x ^{2}} (ln (x))

y^{^{''}} + 4.5 y^{^{'}} = 0

\int x + 4 - \frac{4}{x} d x

What is the solution for y^'=-3y+e^{-2x},y(0)=0 ?
The solution for y^'=-3y+e^{-2x},y(0)=0 is y= 1/(e^{2x)}-1/(e^{3x)}