Browse Questions

# Solve the following $\large\frac{dy}{dx}+\frac{y}{x}=\frac{y^{2}}{x^{2}}$

Toolbox:
• First order , first degree DE
• Variable separable : Variables of a DE are rearranged to separate then, ie
• $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
• Can be written as $\large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
• The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
• A D.E of first order and first degree is said to be homogeneous if it can be put in the form $\large\frac{dy}{dx}=f\bigg(\large\frac{y}{x}\bigg)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)} Where f_1 and f_2 are homogeneous functions in x and y. • To solve we put y=vx and proceed. Step 1: \large\frac{dy}{dx}+\large\frac{y}{x}=\frac{y^2}{x^2} This is a homogeneous DE as \large\frac{dy}{dx}=\frac{y^2}{x^2}-\frac{y}{x} is a function of \large\frac{y}{x} Let y=vx=>\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
Step 2:
$\therefore$ the DE becomes $v+x\large\frac{dv}{dx}=\frac{v^2x^2}{x^2}-\frac{vx}{x}$
$v+x\large\frac{dv}{dx}$$=v^2-v x \large\frac{dv}{dx}$$=v^2-2v$
$\large\frac{dv}{v^2-2v}=\large\frac{dx}{x}$
Step 3:
The variables are separable.
The GS is $\int \large\frac{dv}{v^2-2v}=\int \large\frac{dx}{x}$$+c_1 \int \large\frac{dv}{(v^2-2v+1)-1}=\int \large\frac{dx}{x}$$+c_1$
$I_1=\int \large\frac{dv}{(v^2-2v+1)-1}=\int \large\frac{dv}{(v-1)^2-1}$
$\quad= \large\frac{1}{2}$$\log \bigg(\large\frac{v-1-1}{v-1+1}\bigg) \quad= \large\frac{1}{2}$$ \log \large\frac{v-2}{v}$
Now $v= \large\frac{y}{x}$
$\therefore I_1=\large\frac{1}{2}$$\log \large\frac{\Large\frac{y}{x}-2}{\Large\frac{y}{x}} \quad=\large\frac{1}{2}$$ \log \large\frac{y-2x}{y}$
$\therefore$ the Gs reduces to
$\large\frac{1}{2}$$\log \large\frac{y-2x}{y}$$= \log x -\log c_2$
=> $2 \log c_2=2 \log x -\log \bigg(\large\frac{y-2x}{y}\bigg)$
=>$\log c=\log x^2-\log \bigg(\large\frac{y-2x}{y}\bigg)$
=>$\log c=\log \large\frac{x^2y}{y-2x}$
=>$c(y-2x)=x^2y$