This is not a homework question. I am studying on my own. While basic, yes, I need help applying the general formula. Every time I solve a linear equation it ends up wrong. The answer always shows multiplying by an integrating factor rather than using the formula. Using the formula should get me the same answer.


Putting the equation in general form gives:


where $P(x)= -10$ and $q(x)=10t$

$u(t)= e^{\int-10dt}= e^{-10t}$

the general formula gives $y(t) = \frac{1}{e^{-10t}} \int 10te^{-10t}$

pulling out the constant and integrating by parts gives:

$u=t, du= 1, dv=e^{-10t}, v=\frac{-e^{-10t}}{10}$

which gives

$\frac{-te^{-10t}}{10}-\int -\frac{e^{-10t}}{10}$

which in all equals out to be:

$\frac{10}{e^{-10t}} [\frac{-te^{-10t}}{10}-\frac{e^{-10t}}{100}]$



What is the problem? Please answer using the general formula. NO SHORTCUTS.

  • 2
    $\begingroup$ Are you clarifying that this isn't homework to avoid downvotes? But then won't homework-askers also employ this strategy? $\endgroup$ – Mark Aug 7 '18 at 7:13
  • $\begingroup$ Does the right side of the original equation have ${^-}10t$ or ${^+}10t$? I think you should have $q(t) = -10t$. $\endgroup$ – mr_e_man Aug 7 '18 at 7:32
  • 1
    $\begingroup$ @K.Gibson : You apply a formula without writing what is the formula. You use notations $x,u,v$ without writing where they come from. Really, are you understanding what you are doing ? $\endgroup$ – JJacquelin Aug 7 '18 at 7:33

Your first problem is you have lost a minus sign - re-arranging






First step is to solve the homogeneous equation


and it should be clear that the general solution to this is

$y(t) = ce^{10t}$

for any constant $c$.

So the general solution of $\frac{dy}{dt}-10y=-10t$ is

$y(t) = ce^{10t} + u(t)$

where u(t), the "particular integral", is some specific solution to


Since the right hand side is a polynomial in $t$ this suggests trying a polynomial in $t$ of the same order. So let's try

$u(t) = at+b$

If this is a solution to $\frac{du}{dt}-10u=-10t$ then

$a - 10(at+b) = -10t$

Since this equation must hold for any value of $t$ you can equate the co-efficients of each power of $t$ to find the values of $a$ and $b$.

| cite | improve this answer | |

The last step (“simplified:”) step seems all wrong. You should add “$+c$” already inside the square brackets on the line above, and cancel $e^{-10 t}$ in the second term.

| cite | improve this answer | |

$$\frac{dy}{dt}=10y-10t$$ $$\frac{dy}{dt}-10y=-10t$$ Mulitpl by integrating factor $\mu(t)=e^{-10t}$ $$\frac{dy}{dt}e^{-10t}-10e^{-10t}y=-10te^{-10t}$$ $$(e^{-10t}y)'=-10te^{-10t}$$ Integrate $$\int(e^{-10t}y)'dt=-10\int te^{-10t}dt$$ $$e^{-10t}y+C= te^{-10t}+\frac {e^{-10t}}{10}$$ Finally $$y(t)= t+\frac {1}{10}+Ke^{10t}$$

| cite | improve this answer | |

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.