find_the_fun
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The given family of functions is the general solution of the D.E. on the indicated interval. Find a member of the family that is a solution of the initial-value problem.
[math]y=c_1x+c_2x\ln{x}[/math] on [math](0, \infty)[/math] and [math]x^2y''-xy'+y=0[/math] and y(1)=3, y'(1)=-1
So plugging in y(1)=3 gives [math]3=c_1+c_2\ln{1}[/math] and then take the derivative to get [math]y'=c_1+c_2 \ln{x} +c_2[/math] subbing in [math]-1=C-1+c_2\ln{1}+c_2[/math]
adding 3 times the second equation to the first give [math]0=4c_1+4c_2\ln{1}+3c_2[/math]
What next?
[math]y=c_1x+c_2x\ln{x}[/math] on [math](0, \infty)[/math] and [math]x^2y''-xy'+y=0[/math] and y(1)=3, y'(1)=-1
So plugging in y(1)=3 gives [math]3=c_1+c_2\ln{1}[/math] and then take the derivative to get [math]y'=c_1+c_2 \ln{x} +c_2[/math] subbing in [math]-1=C-1+c_2\ln{1}+c_2[/math]
adding 3 times the second equation to the first give [math]0=4c_1+4c_2\ln{1}+3c_2[/math]
What next?