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# Show that the differential equation is homogeneous and solve $(x^2-y^2)dx+2xy\;dy=0$

Toolbox:
• A differential equation of the form $\large\frac{dy}{dx }$$= F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. • To solve this type of equations substitute y = vx and \large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
By rearranging the equation we get,
$\large\frac{dy}{dx} = -\large\frac{(x^2-y^2)}{2xy }$
Using the information in the tool box let us substitute $y = vx$ and $\large\frac{dy}{x} =$$v + x\large\frac{dv}{dx} v + x\large\frac{dv}{dx} = -\large\frac{(x^2-v^2x^2)}{2xvx} Step 2: Taking x^2 as the common factor and cancelling it on the RHS v + x\large\frac{dv}{dx} = \frac{(1-v^2)}{2v} Bringing v from LHS to the RHS we get, x\large\frac{dv}{dx }= -\frac{(1- v^2) + 2v^2}{2v} x\large\frac{dv}{dx} = -\frac{(1+v^2)}{2v} On seperating the variables we get 2v\large\frac{dv}{(1+v^2) }= -\frac{ dx}{x} Step 3: On integrating both sides we get \int \large\frac{2vdv}{(1+v^2)} = - \int\large\frac{ dx}{x } \log(1+v^2) = -\log x +\log C \log\large\frac{(1+v^2)}{x}$$= \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.