Browse Questions

# Solve the following differential equation : $(x^2-y^2)dx+2xydy=0$ given that y = 1 when x = 1.

Toolbox:
• $y = vx$
• $\large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx} Step 1: (x^2-y^2)dx+2xydy=0 By rearranging the above equation we get, \large\frac{dy}{dx} = -\large\frac{(x^2-y^2)}{2xy } Let us substitute y = vx v + x\large\frac{dv}{dx} = -\large\frac{(x^2-v^2x^2)}{2xvx} Step 2: Taking x^2 as the common factor and cancelling it on the RHS v + x\large\frac{dv}{dx} = \frac{(1-v^2)}{2v} Bringing v from LHS to the RHS we get, x\large\frac{dv}{dx }= -\frac{(1- v^2) + 2v^2}{2v} x\large\frac{dv}{dx} = -\frac{(1+v^2)}{2v} On seperating the variables we get 2v\large\frac{dv}{(1+v^2) }= -\frac{ dx}{x} Step 3: On integrating both sides we get \int \large\frac{2vdv}{(1+v^2)} = - \int\large\frac{ dx}{x } \log(1+v^2) = -\log x +\log C \log\large\frac{(1+v^2)}{x}$$= \log C$
$\large\frac{(1+v^2)}{x} $$= \log C Step 4: Writing the value of v = \large\frac{y}{x} we get \large\frac{[1+(y^2/x^2)]}{x}$$ = C$
$x^2+y^2 = Cx$
This is the required solution.
Step 5:
When $x=1,y=1$
$1+1=C\times 1$
$2=C$
So the required solution is $x^2+y^2=2x$