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Solve the followingdifferential equation:`(x^3+y^3)dy-x^2y dx=0` |
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Answer» `(x^3+y^3) dy - x^2y dx = 0` `=>dy/dx = (x^2y)/(x^3+y^3)` `=>dy/dx = 1/(x/y+y^2/x^2)` Let `y/x = v => y = vx =>dy/dx = v+x(dv)/dx` So, our differential equation becomes, `v+x(dv)/dx = 1/(1/v+v^2) = v/(1+v^3)` `=>x(dv)/dx = v/(1+v^3) - v = -v^4/(1+v^3)` `=> -((1+v^3)/v^4) dv = dx/x` Integrating both sides, `=>int -((1+v^3)/v^4) dv = int dx/x` `=>int -(v^-4+1/v) dv = int dx/x` `=>v^-3/3 - logv = logx +c` `=>1/(3v^3) - c= logx+logv` `=>1/(3v^3) - c= log(vx)` `=>x^3/(3y^3) = logy+c`, which is the required solution. |
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