Solution to the ODEy′(t)=ϕ(t)−α(t)y(t)I encounter the following ODE:∂y(t)∂t=exp⁡(−βt)α−y(t)(γ+exp⁡(−βt))with α,β and γ real scalars. and y(0)=1.0How to solve it in order to get y(t) ?

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Solution to the ODEy′(t)=ϕ(t)−α(t)y(t)I encounter the following ODE:∂y(t)∂t=exp⁡(−βt)α−y(t)(γ+exp⁡(−βt))with α,β and γ real scalars. and y(0)=1.0How to solve it in order to get y(t) ?

Annie Levine · Accepted Answer

Step 1You need to use an Integrating factor which I denote as I(t) which you multiply inIy′=Iϕ(t)−Iα(t)yBring all the y dependent terms to one sideIy′+Iα(t)y=Iϕ(t)the left is turned into product rule of differentiationso d(Iy)dt=Iϕ(t)Step 2I is now the followingdIdt=Iα(t)so I=const⋅e∫α(t)dt which for α scalar isI=ceαt

ophelialee4xn · Accepted Answer

Step 1To start with, rewrite the equation in standard form,y′(t)+[γ+exp⁡(−β{t})]y(t)=αexp⁡(−β{t}).Let&amp;nbsp;c(t)=γ+exp⁡(−β{t}), soy′(t)+c(t)y(t)=αexp⁡(−β{t}),and let&amp;nbsp;I(t)=exp⁡(∫0tc(t&#039;),dt&#039;).You can prove thatI&#039;(t)=c(t)exp⁡(∫0tc(t&#039;),dt&#039;)=c(t)I(t).Therefore, multiplying the differential equation by I(t) results inI(t)y′(t)+c(t)I(t)y(t)=I(t)y′(t)+I′(t)y(t)=(I⋅{y})′(t)=I(t)αexp⁡(−β{t}).Therefore,&amp;nbsp;I(t)y(t)−I(0)y(0)=α∫0tI(t&#039;)exp⁡(−β{t&#039;}),dt&#039;.Step 2Now&amp;nbsp;I(0)y(0)=1, given the initial conditions. However, to proceed further, we need to find a simpler formula for I(t). notice that∫0tγ+exp⁡(−βt&#039;)&amp;nbsp;dt&#039;={γt+1−exp⁡(−βt)ββ≠0(γ+1)tβ=0soI(t)={exp⁡(1β)exp⁡(γt)exp⁡(−exp⁡(−βt)β)β≠0exp⁡((γ+1)t)β=0.In the case that&amp;nbsp;β=0, this means thatα∫0tI(t&#039;)exp⁡(−βt&#039;)&amp;nbsp;dt&#039;={αtγ=−1α1−exp⁡((γ+1)t)γ+1γ≠−1,implying thaty(t)={(1+αt)exp⁡(−(γ+1)t)γ=−1exp⁡(−(γ+1)t)+αγ+1exp⁡(−(γ+1)t)−1γ+1γ≠−1.Naturally, the case where&amp;nbsp;β≠0&amp;nbsp;is much trickier, asα∫0tI(t&#039;)exp⁡(−β{t&#039;}),dt&#039;=αexp⁡(1β)∫0texp⁡(γ{t&#039;})exp⁡(−exp⁡(−β{t&#039;})β)exp⁡(−β{t&#039;}),dt&#039;.Notice that if we let&amp;nbsp;s=−exp⁡(−β{t})β, then&amp;nbsp;dsdt=exp⁡(−β{t}), and&amp;nbsp;−β{s}=exp⁡(−β{t}), implying&amp;nbsp;ln⁡(−β{s})=−β{t}, so&amp;nbsp;t=−ln⁡(−β{s})β, and thus,&amp;nbsp;exp⁡(γ{t})=exp⁡(−γβln⁡(−β{s})). To explore further, if&amp;nbsp;β&amp;lt;0, thenexp⁡(−γβln⁡(−β{s}))=exp⁡(−γβln⁡(−β))exp⁡(−γβln⁡(s))=(−β)−γβs−γβ,hence&amp;nbsp;αexp⁡(1β)∫0texp⁡(γ{t&#039;})exp⁡(−exp⁡(−β{t&#039;})β)exp⁡(−β{t&#039;}),dt&#039;=α(−β)−γβexp⁡(1β)∫−1β−exp⁡(−β{t})βs−γβexp⁡(s),ds&#039;.The analysis for&amp;nbsp;β&amp;gt;0&amp;nbsp;proceeds similarly.

Solution to the ODE y'(t) =\phi(t) - \alpha(t) y(t) I encounter the

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