What is the solution for 6y^'-y=9x^{-3x},y(0)=4 ?

The solution for 6y^'-y=9x^{-3x},y(0)=4 is y=e^{x/6}*integral of (3e^{-x/6}x^{-3x})/2

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6y^'-y=9x^{-3x},y(0)=4

6 y^{^{'}} - y = 9 x^{- 3 x}, y (0) = 4

y = e^{\frac{x}{6}} \cdot \int \frac{3 e ^{- \frac{x}{6}} x ^{- 3 x}}{2} d x

Solution steps

6 y^{'} (x) - y = 9 x^{- 3 x}

Solve linear ODE:

y = e^{\frac{x}{6}} \cdot \int \frac{3 e ^{- \frac{x}{6}} x ^{- 3 x}}{2} d x

y = e^{\frac{x}{6}} \cdot \int \frac{3 e ^{- \frac{x}{6}} x ^{- 3 x}}{2} d x

\frac{2 x + 4}{3 x + 7}, at - 2

\frac{d}{d x} (\frac{sin ( x y )}{e ^{x} - y ^{2}})

2 x y^{2}

n = 0 \sum \infty n \cdot (\frac{2}{3})^{n}

4 e^{2 x^{2} - 4}

What is the solution for 6y^'-y=9x^{-3x},y(0)=4 ?
The solution for 6y^'-y=9x^{-3x},y(0)=4 is y=e^{x/6}*integral of (3e^{-x/6}x^{-3x})/2