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Assignment Statement

given $$ y^{\prime} + p(t)y = q(t) $$ a general linear equation where $q(t)$ is not zero everywhere,

I'm asked to find $C(t)$ form $$ C^{\prime}(t)= q(t)e^{\int p(t) \ dt}$$ then substitute in $$ y(t) = C(t)e^{-\int p(t) \ dt}$$ to extract the y(t) solution.

Attempt at solution

I'm having trouble at the first step if I integrate both sides I get $$ C(t) = \int\left( q(t) e^{\int p(t) \ dt}\right) \ dt$$

I feel like I have to do integration by parts on the right-hand side but both substitutions give me very odd expressions involving integrals of integrals and I'm genuinely confused how to start.

I also tough that maybe I just substitute everything and do something on the final expression which is: $$ y(t) = \frac{\int\left( q(t) e^{\int p(t) \ dt}\right) \ dt}{e^{\int p(t) \ dt}}$$ I notice there's a common factor $e^{\int p(t) \ dt}$ maybe I have to do something with that ?

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When you apply the method of $\textrm{variation of parameters}$ you firstly have to solve the homogeneous equation $$ y^{\prime} + p(t)\cdot y =0 $$

$$y^{\prime} =- p(t)\cdot y $$

$$\frac{dt}{y}=-p(t) \, dt$$

$$ \ln(y)=-\int p(t) \, dt$$

$$y_h=C\cdot e^{-\int p(t) \, dt}$$

Then the non-homogeneous solution is $y_p=C(t)\cdot e^{-\int p(t) \, dt}$. Differentiating the function gives

$y'_p=C'(t)\cdot e^{-\int p(t) \, dt}-C(t)\cdot p(t) \cdot e^{-\int p(t) \, dt}$.

Plugging into the non-homogeneous equation.

$C'(t)\cdot e^{-\int p(t) \, dt}\underbrace{-C(t)\cdot p(t) \cdot e^{-\int p(t) \, dt}+p(t)\cdot C(t)\cdot e^{-\int p(t) \, dt}}_{=0}=q(t)$

$C'(t)\cdot e^{-\int p(t) \, dt}=q(t)$

$C'(t)=q(t)\cdot e^{\int p(t) \, dt}\Rightarrow C(t)=\int q(t)\cdot e^{\int p(t) \, dt} \, dt$. Using this result to obtain $y_p$.

$y_p=e^{-\int p(t) \, dt} \cdot \int q(t)\cdot e^{\int p(t) \, dt} \, dt$. It total we get

$$y=y_h+y_p=C\cdot e^{-\int p(t) \, dt}+e^{-\int p(t) \, dt} \cdot \int q(t)\cdot e^{\int p(t) \, dt} \, dt$$

$$=\left(C+\int q(t)\cdot e^{\int p(t) \, dt} \, dt\right)\cdot e^{-\int p(t) \, dt}$$

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  • $\begingroup$ Thank you for your reply this seems right, I just don't quite understand the step : $$ y = y_{h} + y_{p}$$ Does it say that the solution to $y$ is the sum of the homogeneous with the non-homogeneous solution ? Is that like a theorem or something , because I don't remember seen this in class. $\endgroup$
    – hexaquark
    Jan 24 '20 at 13:30
  • $\begingroup$ @hitechphysics Yes, it is the sum. It is hard to explain in short. Maybe the first pages of this document make it clearer. $\endgroup$ Jan 24 '20 at 14:41
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Multiply both sides of $$y^{\prime} + p(t)y = q(t)$$ by $$ e^{\int p(t)dt} $$ to get $$y^{\prime}e^{\int p(t)dt} + p(t)e^{\int p(t)dt}y = e^{\int p(t) dt}q(t)$$

The LHS is the derivative of $$ye^{\int p(t)dt}$$ Thus upon integration of both sides you get $$ye^{\int p(t)dt} = \int e^{\int p(t) dt}q(t) +C$$

Solve for $y$ and you get $$y =e^{-\int p(t)dt} \int e^{\int p(t) dt}q(t) \ dt+Ce^{-\int p(t)dt}$$

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  • $\begingroup$ Thank you for your reply, but does this solving method actually use the method of variation of parameters ? $\endgroup$
    – hexaquark
    Jan 23 '20 at 17:45
  • $\begingroup$ This method is called the integrating factor method but as you mentioned it is a variation of parameters method. $\endgroup$ Jan 23 '20 at 19:47

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