Question

(1/x)(dy/dx)=((ysinx)/(y^2+1))

initial value problem?

Answer 1

\[\frac{ 1 }{ x }*\frac{ dy }{ dx }=\frac{ ysinx }{ y ^{2} +1}\] 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

Answer 2

thanks once again!!!

Answer 3

Yeah, sure :3