Differential equation substitution method

[tony873004]

$$\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{4x^2 + 5xy + y^2 }}{{x^2 }} \\ \\ \frac{{dy}}{{dx}} = 4 + \frac{{5y}}{x} + \left( {\frac{y}{x}} \right)^2 \\ \\ {\rm{Let }}v = y/x\,\,\,\, \Rightarrow \,\,\,\,y = vx \\ \\ \frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{xx}} + \left( {\frac{{vx}}{x}} \right)^2 \\ \\ \frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{x^2 }} + v^2 \\ \end{array}$$

But the class notes say the final line should be
$$\frac{dy}{dx}=x\frac{dv}{dx}+v$$
How did he get that?

 

[waht]

Take the derivative of y = vx with respect to x, then equate it, and you are done.

 

[gabbagabbahey]

$$\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{4x^2 + 5xy + y^2 }}{{x^2 }} \\ \\ \frac{{dy}}{{dx}} = 4 + \frac{{5y}}{x} + \left( {\frac{y}{x}} \right)^2 \\ \\ {\rm{Let }}v = y/x\,\,\,\, \Rightarrow \,\,\,\,y = vx \\ \\ \frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{xx}} + \left( {\frac{{vx}}{x}} \right)^2 \\ \\ \end{array}$$

This should be $\frac{dy}{dx}= 4 + \frac{{5vx}}{{x}} + \left( {\frac{{vx}}{x}} \right)^2=4+5v+v^2$

But the class notes say the final line should be
$$\frac{dy}{dx}=x\frac{dv}{dx}+v$$
How did he get that?

Use the product rule to differentiate $y=vx$.