What is the easiest way to solve the following differential equation? I am only just starting online courses on the subject of differential equations, and I passed the following differential equation
$$\frac{dy}{dx}= -2x + 3y - 5$$
I'm a bit confused by the use of two variables in the equation. What is the easiest way to solve this? In other words, in a way that a complete differential equation beginner like me can understand it.
 A: $$\frac{dy}{dx}= -2x + 3y - 5$$
$$\frac{dy}{dx}+3y= -2x-5$$
$$\text{Integrating Factor}: \mu=e^{\int -3dx} =e^{-3x}$$
Multiplying through by $\mu$
$$ \frac{dy}{dx}e^{-3x}+3e^{-3x}y = -2xe^{-3x}-5e^{-3x}$$
Can you continue?
A: An alternative method would be to use the following substitution to reduce the first-order linear ODE into a separable one:
$$v=-2x+3y\implies y=\frac{v+2x}{3}\implies \frac{dy}{dx}=\frac{1}{3}\left(\frac{dv}{dx}+2\right)$$

Therefore, we have:
$$\frac{1}{3}\left(\frac{dv}{dx}+2\right)=v-5$$
$$\frac{dv}{dx}=3v-17$$
This can now be separated to give:
$$\int \frac{dv}{3v-17}=\int dx$$
Can you continue? After you solve for $v(x)$ explicitly, be sure to substitute back to obtain the general solution for $y(x)$.
A: We can solve equation by the Variation of parameters
$$\frac { dy }{ dx } =-2x+3y-5\\ \frac { dy }{ dx } -3y=0\\ \int { \frac { dy }{ y }  } =3\int { dx } \\ \ln { y=3x+C } \\ y=C{ e }^{ 3x }\\ y=C\left( x \right) { e }^{ 3x }\\ { y }^{ \prime  }={ C }^{ \prime  }\left( x \right) { e }^{ 3x }+3{ C\left( x \right) e }^{ 3x }\\ { C }^{ \prime  }\left( x \right) { e }^{ 3x }+3{ C\left( x \right) e }^{ 3x }=-2x+3C\left( x \right) { e }^{ 3x }-5\\ { C }^{ \prime  }\left( x \right) { e }^{ 3x }=-2x-5\\ { C }^{ \prime  }\left( x \right) ={ e }^{ -3x }\left( -2x-5 \right) \\ { C }\left( x \right) =-2\int { { xe }^{ -3x } } dx-5\int { { e }^{ -3x }dx } +C=-2\left[ -\frac { 1 }{ 3 } \int { xd\left( { e }^{ -3x } \right)  }  \right] +\frac { 5 }{ 3 } { e }^{ -3x }+C=\\ =\frac { 2 }{ 3 } \left[ x{ e }^{ -3x }-\int { { e }^{ -3x }dx }  \right] +\frac { 5 }{ 3 } { e }^{ -3x }+C=\frac { 2 }{ 3 } \left[ x{ e }^{ -3x }+\frac { { e }^{ -3x } }{ 3 }  \right] +\frac { 5{ e }^{ -3x } }{ 3 } ={ e }^{ -3x }\left[ \frac { 2x }{ 3 } +2 \right] +C\\ y=C{ \left( x \right) e }^{ 3x }=\frac { 2x }{ 3 } +2+C$$
so

$$ \color{blue}\ {y=\frac { 2x }{ 3 } +2} $$

