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# First Order Differentials watch

1. I have a question;

Find the general solution to the differential eqtn. dy/dx = (x^2+y^2)/(2xy), using the substitution y=xu

My attempt:

Using y =xu,

dy/dx = u+xdu/dx

Then, we make this equal to the diff. eqtn on the top of the page;

u+xdu/dx = (x^2+y^2)/(2xy)

Now, we substiture y=xu into this;

u+xdu/dx = (x^2+x^2.u^2)/(2x^2.u)

Now, on the RHS, we can cancel x^2 terms;

u+xdu/dx = (1+uu)/2u (nb uu=u^2) (!)

then we bring the u over;

xdu/dx = (1+uu-2uu)/2u

xdu/dx = (1-uu)/(2u)

Then i tried to separate variables:

(2u)/(1-uu) du = 1/x dx

which i then integrate, to give -ln|1-uu| = lnx + K, a constant

but the correct answer is;
yy=x(x+k)

Where am I going wrong, please?
2. From -ln|1-uu| = lnx + K, you have exp(-ln(1-uu)) = exp(ln x)exp(K)

Then rewrite u = y/x and it should come out.

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Updated: March 30, 2011
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