<s>[INST] 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.
 [/INST] 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

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

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