Question

Please help me solving this question. Find dy/dx: xe^2y=5x+y^2.

Answer 1

Since you can't separate the variables ie. cleanly have y on one side with all non-y stuff on the other then do as follows :\[xe^{2y}=5x+y^{2}\]take derivative of both sides with respect to x :\[\frac{d}{dx}(xe^{2y})=\frac{d}{dx}(5x+y^{2})\]apply product rule on LHS, linearity of derivative on RHS :\[x\frac{d}{dx}(e^{2y})+e^{2y}\frac{d}{dx}(x)=\frac{d}{dx}(5x)+\frac{d}{dx}(y^{2})\]derivative of x with respect to itself is 1, derivative of a constant times x is that constant :\[x\frac{d}{dx}(e^{2y})+e^{2y}=5+\frac{d}{dx}(y^{2})\]chain rule :\[x\frac{d}{dy}(e^{2y})\frac{dy}{dx}+e^{2y}=5+\frac{d}{dy}(y^{2})\frac{dy}{dx}\]derivative of exponential with linear exponent is itself ( care with constant multiplier in exponent ), derivative of power is to reduce power and multiply by previous power\[x(2e^{2y})\frac{dy}{dx}+e^{2y}=5+2y\frac{dy}{dx}\]re-arrange\[\frac{dy}{dx}(2xe^{2y}-2y)=5-e^{2y}\]\[\frac{dy}{dx}=\frac{5-e^{2y}}{2xe^{2y}-2y}=\frac{5-e^{2y}}{2(xe^{2y}-y)}\]horrible, but that's as simple as you'll get it.