InterviewSolution
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InterviewSolution
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`[ x tan(y/x) - y sec^2(y/x) ] dx + x sec^2 (y/x)] dy= 0` |
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Answer» Let `y = vx` Then, `dy/dx = v+x(dv)/dx` Our given equation becomes, `[xtanv-vxsec^2 v]dx + xsec^2v dy = 0` `=>-dy/dx = [xtanv-vxsec^2 v]/[xsec^2v]` `=>-v-x(dv)/dx = tanv/(sec^2v) - v` `=>- dx/x = (sec^2v dv)/tanv` Now, integrating both sides, `=> int - dx/x = int (sec^2v dv)/tanv` Let `tanv = u => sec^2vdv = du` `=> int - dx/x = int (du)/u` `=>-ln(x)+c = ln(u)` `=>c = ln(ux) ` `=>ln(xtanv) = c` `=>ln(xtan(y/x)) = c` |
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