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# Solve : $(x^2+xy)dy=(x^2+y^2)dx$

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## 1 Answer

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Toolbox:
• A function is said to be homogenous function in degree n if $F(kx,ky) = k^nF(x,y)$ for any non zero constant $k.$
• To solve these type of homogenous functions we make the substitution $y = vx$ hence $\large\frac{dy}{dx}$$= v+x\large\frac{dv}{dx} Step 1: Given: (x^2 + xy)dy = (x^2 + y^2)dx \large\frac{dy}{dx} =\frac{ (x^2 + y^2)}{(x^2 + xy)} F(x,y) =\large\frac{ (x^2 + y^2)}{(x^2+xy)} F(kx,ky) = \large\frac{k[x^2+y^2]}{[x^2 + xy]}$$ = k^0$
Step 2:
Using the information in the tool box let us substitute for $y = vx$
$\large\frac{dy}{dx}$$= v + x\large\frac{dv}{dx} v + x\large\frac{dv}{dx} =\frac{ [x^2 +x^2 v^2]}{x^2 + xvx} Taking the common factor x^2 and cancelling we get v + x\large\frac{dv}{dx} =\frac{ [1+v^2]}{1+v} bringing v from the LHS to the RHS and simplifying we get, x\large\frac{dv}{dx} =\frac{ 1-v}{1+v} on seperating the variables we get, \large\frac{[1+v]}{[1-v]} =\frac{ dx}{x} \large\frac{2}{[1-v] - 1} =\frac{ dx}{x} Step 3: On integrating both sides we get, -2\log[1-v] -v = \log x + \log C v = -2\log[1-v] - \log x +\log C v=\large\frac{\log c}{x(1-v)^2} substituting for v we get \large\frac{y}{x} =\frac{ \log C}{x[1-(y/x)^2]} \large\frac{cx}{(x-y)^2 }=$$e^{\Large (y/x)}$
on rearranging we get

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