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Solve the following differential equation `x^2(dy)/(dx)-xy=1+cos(y/x), x!=0` |
Answer» `x^2dy/dx - xy = 1+cos(y/x)` `=>dy/dx -y/x = 1/x^2(1+cos(y/x))` Let `y = vx`, then `dy/dx = v+x(dv)/dx` Now, our equation becomes, `v+x(dv)/dx -v = 1/x^2(1+cosv)` `=>(dv)/(1+cosv) = 1/x^3dx` `=>1/2sec^2(v/2) dv = dx/x^3` Integrating both sides, `=>1/2 int sec^2(v/2) dv = int dx/x^3` `=>1/2(2)tan(v/2) = -1/2 x^-2+c` `=>tan(v/2) = -1/(2x^2)+c` `=>tan(y/(2x)) = -1/(2x^2)+c` |
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