Answer to Question #259057 in Differential Equations for jayzie

Question

(10-6y+e^(-3x))dx-2dy=0 

Answer 1

$$(10-6y+e^{-3x})dx-2dy=0$$

$$y'+3y=\dfrac{1}{2}e^{-3x}+5$$

Find the integration factor

$$\mu=e^{3x}$$

Then

$$y'e^{3x}+3ye^{3x}=\dfrac{1}{2}+5e^{3x}$$

$$d(e^{3x}y)=(\dfrac{1}{2}+5e^{3x})dx$$

Integrate

$$\int d(e^{3x}y)=\int(\dfrac{1}{2}+5e^{3x})dx$$

$$e^{3x}y=\dfrac{1}{2}x+\dfrac{5}{3}e^{3x}+C$$

$$y=\dfrac{1}{2}xe^{-3x}+\dfrac{5}{3}+Ce^{-3x}$$