What is the solution for xy^'+3y=3e^x ?

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

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xy^'+3y=3e^x

x y^{^{'}} + 3 y = 3 e^{x}

y = \frac{3 ( x ^{2} e ^{x} - 2 ( e ^{x} x - e ^{x} ) )}{x ^{3}} + \frac{c _{1}}{x ^{3}}

Solution steps

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

Solve linear ODE:

y = \frac{3 ( x ^{2} e ^{x} - 2 ( e ^{x} x - e ^{x} ) )}{x ^{3}} + \frac{c _{1}}{x ^{3}}

y = \frac{3 ( x ^{2} e ^{x} - 2 ( e ^{x} x - e ^{x} ) )}{x ^{3}} + \frac{c _{1}}{x ^{3}}

(1 - 5 y + x) \frac{d y}{d x} + y = 5 x - 1

y^{'} (t) = 10 - e^{y}, y (0) = 1

y = x^{3} - \frac{x ^{2}}{16}

\int 5 x + y d y

\frac{d}{d x} (- 2 x)

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