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 ?


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

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

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

  • $\begingroup$ 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

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 $$y =e^{-\int p(t)dt} \int e^{\int p(t) dt}q(t) \ dt+Ce^{-\int p(t)dt}$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.