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Worked solution

Recognise an exact product derivative

The variables satisfy xeydydx+ey=2xxe^y\dfrac{dy}{dx}+e^y=2x.

(i) Use z=xeyz=xe^y to obtain an equation relating zz and xx.

(ii) Hence, or otherwise, show that the general solution is y=ln⁡(x+k/x)y=\ln(x+k/x), where kk is arbitrary.

[2007/MJC/MYE/II/4(b)]

CJC · Discussion · Question 3

Answer

(i) dz/dx=2xdz/dx=2x. (ii) y=ln⁡(x+k/x)y=\ln(x+k/x), with x≠0x\ne0 and x+k/x>0x+k/x>0.

(i) Differentiate the substituted product

z=xeyz=xe^y gives dz/dx=ey+xeyy′=2xdz/dx=e^y+xe^yy^\prime=2x.

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(ii) Integrate and return to y

z=x2+Cz=x^2+C, so xey=x2+Cxe^y=x^2+C. Thus ey=x+C/xe^y=x+C/x and y=ln⁡(x+k/x)y=\ln(x+k/x) with k=Ck=C. Require x≠0x\ne0 and x+k/x>0x+k/x>0. Indeed at x=0x=0 the original equation would demand ey=0e^y=0, which has no finite real solution.

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Common mistakes

Keep the positivity requirement when taking a real logarithm.

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