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#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

\(\displaystyle y=c_1x+c_2x\ln{x}\) on \(\displaystyle (0, \infty)\) and \(\displaystyle x^2y''-xy'+y=0\) and y(1)=3, y'(1)=-1

So plugging in y(1)=3 gives \(\displaystyle 3=c_1+c_2\ln{1}\) and then take the derivative to get \(\displaystyle y'=c_1+c_2 \ln{x} +c_2\) subbing in \(\displaystyle -1=C-1+c_2\ln{1}+c_2\)

adding 3 times the second equation to the first give \(\displaystyle 0=4c_1+4c_2\ln{1}+3c_2\)

What next?