anonymous
  • anonymous
(1/x)(dy/dx)=((ysinx)/(y^2+1)) initial value problem?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Psymon
  • Psymon
\[\frac{ 1 }{ x }*\frac{ dy }{ dx }=\frac{ ysinx }{ y ^{2} +1}\] So let's get our variables separated first: \[\frac{ y ^{2}+1 }{ y }*dy=xsinxdx\] So now we can integrate both sides. The left side can be done by separating into two fractions while the right side can be done using integration by parts. \[\int\limits_{}^{}\frac{ y ^{2} }{ y }dy+\int\limits_{}^{}\frac{ 1 }{ y }=\int\limits_{}^{}xsinxdx\] Now the left two we can do easily, so let's get the right side done. In case we need to recall, this is the formula for ibp: \[u \int\limits_{}^{}v-[\int\limits_{}^{}u'\int\limits_{}^{}v]\]So we choose our u to be x in this case and v to be sinx. Meaning u' is just 1 and integral v is -cosx. So plugging in those values we get: \[-xcosx + \int\limits_{}^{}cosxdx \]I simply factored out the negative at the end part. So finishing that integration I have: \[-xcosx + sinx+C\] So that is the right side. Now the left aide was easy, so we can just fill that in: \[\frac{ y ^{2} }{ 2 }+\ln|y|=-cosx+sinx + C \]Now just apply your initial value and solve for C
anonymous
  • anonymous
thanks once again!!!
Psymon
  • Psymon
Yeah, sure :3

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