# Thread: Solve an IVP and give the largest interval

1. ## Solve an IVP and give the largest interval

I really need help with this problem, can someone show me how to work it out step by step if possible.

Thank You so much

Solve an IVP and give the largest interval I over which the solution is defined

(x+1) dy/dx + y = ln x ; y(1) = 10

2. Originally Posted by kithy
I really need help with this problem, can someone show me how to work it out step by step if possible.

Thank You so much

Solve an IVP and give the largest interval I over which the solution is defined

(x+1) dy/dx + y = ln x ; y(1) = 10

If you put this in standard from and get the integrating factor you will find that it is (x+1). It will turn back into what you started with we notice that the left hand side is the the implicit derivative $\frac{d}{dx}[(x+1)y]$

So we get

$\frac{d}{dx}[(x+1)y]=\ln|x|$ so now we integrate both sides to get

$(x+1)y=x\ln|x|-\ln|x|+c \iff y=\frac{x\ln(x)}{(x+1)}-\frac{\ln(x)}{(x+1)}+\frac{c}{(x+1)}$

using the intial condition we get

$10=\frac{1\ln(1)}{(1+1)}-\frac{\ln(1)}{(1+1)}+\frac{c}{(1+1)} \iff 20 =c$

So we get

$y=\frac{x\ln(x)}{(x+1)}-\frac{\ln(x)}{(x+1)}+\frac{20}{(x+1)}$

For the interval It depends on if you had $\ln(x) \mbox{ or } \ln|x|$

If it is the first then solution only exits for $x \in (x,\infty)$

If it is the 2nd the only singular point is x=-1 so the two intervals would be

$(-\infty,-1) \And (-1,\infty)$

So the largest would be $x \in (-1,\infty)$