Question

Solve the IVP ty′ + (t + 1)y = t, y(ln 2) = 1 for t > 0

Answer 1

Divide both sides by t first:
\[y'+\frac{t+1}{t}y=1\]
now we should choose a v to multiply on both sides so that we can express the left hand side as vy'+v'y so we can write (vy)'

so we need \[v'=\frac{t+1}{t}v\]

but we can find this v by using separation of variables

\[\frac{dv}{dt}=\frac{t+1}{t}v\]

\[\frac{1}{v}dv=(1+\frac{1}{t}) dt\]

now integrate both sides giving us

\[lnv=t+lnt+C\]

we don't need the constant (let C=0)

\[lnv=t+lnt\]

\[v=e^{t+lnt}\]

so we multiply both sides by our v giving us:

\[e^{t+lnt}y'+e^{t+lnt}*\frac{t+1}{t}y=e^{t+lnt}\]

so now we can write:

\[(e^{t+lnt}y)'=e^{t+lnt}\]

now we integrate both sides because we know this will bring us closer to solving it:

\[e^{t+lnt}y=\int\limits_{}^{}e^{t+lnt}dt+K\]

but \[\int\limits_{}^{}e^{t+lnt}dt=\int\limits_{}^{}e^te^{lnt}dt=\int\limits_{}^{}te^tdt=te^t-\int\limits_{}^{}e^tdt=te^t-e^t+K_1\]

so we have

\[e^{t+lnt}y=te^t-e^t+K\]

solving for y gives:

\[y=\frac{te^t-e^t+K}{e^{t+lnt}}\]

Answer 2

and that was for v>0 of course
also you have initial condition
y(ln2)=1

so just plug in ln2 and set =1 to solve for the constant K

\[1=y(\ln2)=\frac{\ln2*e^{\ln2}-e^{\ln2}+K}{e^{\ln2+\ln(\ln2)}}\]

Answer 3

I got y=1-1/t+2/te^t treating it like a linear equation.

Answer 4

yes thats right we can simplify what i have gj dave

Answer 5

Myininaya, can you help me with a de problem?

Answer 6

http://openstudy.com/groups/mathematics/updates/4e27bd7a0b8b3d38d3b8b705

Answer 7

\[y=\frac{te^t-e^t+k}{e^te^{lnt}}=\frac{t-1}{e^{lnt}}+\frac{K}{te^t}=\frac{t-1}{t}+\frac{K}{te^t}\]
\[=1-\frac{1}{t}+\frac{K}{te^t}\]

Answer 8

wendy are you there?
i went the long way to solve it
but there is a short way using a formula

Answer 9

yes! I understand what you did to solve it. Thank you so much!!

Answer 10

np