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Mathematics 7 Online
OpenStudy (anonymous):

Consider the differential equation: \[x \frac{ dy }{ dx }-2y=0\] Find a general solution to this differential equation that has the form y=Cx^n. I wasn't sure how to do this. First I separated the terms (forget what thats called) so I had: \[\frac{ dy }{ 2y }=\frac{ dx }{ x }\] and integrated both sides giving me: \[\frac{ 1 }{ 2 }\ln(y)+c = \ln(x)+c\] but I am stuck as to what to do next. Am I doing something wrong so far?

OpenStudy (anonymous):

By exponentiating both sides, I get y=2x+c, which doesn't seem to be correct.

OpenStudy (anonymous):

|dw:1352445714021:dw|

OpenStudy (anonymous):

integrating both sides (1/2)ln(y)=ln(x)+ln(c) ln(y)=(1/2)ln(x)+(1/2)ln(c) y=e^((1/2)ln(x))+e^((1/2)ln(c))

OpenStudy (anonymous):

|dw:1352445781916:dw|

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