Solve the following differential equation 7

Question

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-7x+3y+7}{3x-7y-3}$$

Do I need to use substitution? There's nothing like this in the book, so I'm confused.

Thanks in advance.

Answer 1

If we set $x=t+1$, our DE takes the simpler form $$ \frac{dy}{dt} = \frac{-7t+3y}{3t-7y} \tag{1} $$ and by setting $y(t)=\frac{3}{7}t+u(t)$ the previous identity turns into $$ 24+7\,u'(t) = \frac{40\,t}{u(t)}\tag{2} $$ then by setting $u(t)=t\cdot v(t)$ we get $u'(t)=v(t)+t\cdot v'(t)$ and $$ \frac{t\,dv}{dt} = \frac{40}{v}-24-v\tag{3} $$ or: $$ \frac{dt}{t} = \frac{dv}{\frac{40}{v}-24-v}\tag{4} $$ that finally is a separable differential equation, giving $t$ in terms of $v$.
To solve $(1)$ it is enough to perform a function inversion and two simple substitutions.