Differential equation substitution method

tony873004
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\begin{array}{l}<br /> \frac{{dy}}{{dx}} = \frac{{4x^2 + 5xy + y^2 }}{{x^2 }} \\ <br /> \\ <br /> \frac{{dy}}{{dx}} = 4 + \frac{{5y}}{x} + \left( {\frac{y}{x}} \right)^2 \\ <br /> \\ <br /> {\rm{Let }}v = y/x\,\,\,\, \Rightarrow \,\,\,\,y = vx \\ <br /> \\ <br /> \frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{xx}} + \left( {\frac{{vx}}{x}} \right)^2 \\ <br /> \\ <br /> \frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{x^2 }} + v^2 \\ <br /> \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?
 
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Take the derivative of y = vx with respect to x, then equate it, and you are done.
 
tony873004 said:
\begin{array}{l}<br /> \frac{{dy}}{{dx}} = \frac{{4x^2 + 5xy + y^2 }}{{x^2 }} \\ <br /> \\ <br /> \frac{{dy}}{{dx}} = 4 + \frac{{5y}}{x} + \left( {\frac{y}{x}} \right)^2 \\ <br /> \\ <br /> {\rm{Let }}v = y/x\,\,\,\, \Rightarrow \,\,\,\,y = vx \\ <br /> \\ <br /> \frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{xx}} + \left( {\frac{{vx}}{x}} \right)^2 \\ <br /> \\ <br /> \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.
 
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