Solving $\frac{dy}{dx}=\sqrt{3x+2y}-\frac{3}{2}$ without stuff from higher-order differential equations

I'm trying to solve this equation: $$\frac{dy}{dx}=\sqrt{3x+2y}-\frac{3}{2}$$ without using stuff from higher-order differential equations.

I've tried using substitution $$w=\frac{y}{x}$$, but it doesn't really help. After using that substitution, I get $$\frac{dw}{dx}=\frac{\sqrt{x}\sqrt{3+2w}-\frac{3}{2}w}{x}$$ which isn't a differential equation with separable variables.

• Which stuff ? – Yves Daoust Apr 22 at 13:14

Hint: Substitute $$u = 3x + 2y$$. It should then be separable.
Rearranging and squaring gives $$\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}+\frac94=3x+2y$$ Differentiating gives that $$2\cdot \frac{dy}{dx}\frac{d^2y}{dx^2}+3\cdot\frac{d^2y}{dx^2}=3+2\cdot\frac{dy}{dx}$$ $$\frac{d^2y}{dx^2}\left(2\cdot\frac{dy}{dx}+3\right)=2\cdot\frac{dy}{dx}+3$$ $$\frac{d^2y}{dx^2}=1$$ $$y=\frac12x^2+C_1x+C_2$$ Plugging this in the original equation we have $$x+C_1=\sqrt{x^2+(2C_1+3)x+2C_2}-\frac32$$ $$\left(x+C_1+\frac32\right)^2=x^2+(2C_1+3)x+2C_2$$ $$x^2+\left(2C_1+3\right)x+\left(C_1+\frac32\right)^2=x^2+(3+2C_1)x+2C_2$$ $$\therefore C_2=\frac12\left(C_1+\frac32\right)^2$$ Hence the general solution is $$y=\frac12x^2+C_1x+\frac12\left(C_1+\frac32\right)^2$$
• You can get the results cheaper. For $y''=1$, insert $y'=x+c$ into the squared equation, for the second case insert $y=-\frac32$ into the first order equation. – LutzL Apr 22 at 13:10