What is the solution for xy^'+(2x-3)y=4x^4 ?

The solution for xy^'+(2x-3)y=4x^4 is y=2x^3+(c_{1}x^3)/(e^{2x+c_{1)}}

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xy^'+(2x-3)y=4x^4

x y^{^{'}} + (2 x - 3) y = 4 x^{4}

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

Solution steps

x y^{'} (x) + (2 x - 3) y = 4 x^{4}

Solve linear ODE:

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

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

\frac{( s + 1 )}{( s + 2 ) ( s - 3 )}

4 y - y^{^{''}} = 10 e^{- 3 x}

\int ln (x) + 1 d x

\int 2 sin (x) + cos (x) d x

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

What is the solution for xy^'+(2x-3)y=4x^4 ?
The solution for xy^'+(2x-3)y=4x^4 is y=2x^3+(c_{1}x^3)/(e^{2x+c_{1)}}