# First order linear differential equation. Using the general solution

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.

$\frac{dy}{dt}=10y-10t$

Putting the equation in general form gives:

$\frac{dy}{dt}-10y=10t$

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:

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

simplified:

$-t-\frac{e^{-10t}}{10}+c$

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

• Are you clarifying that this isn't homework to avoid downvotes? But then won't homework-askers also employ this strategy? – Mark Aug 7 '18 at 7:13
• Does the right side of the original equation have ${^-}10t$ or ${^+}10t$? I think you should have $q(t) = -10t$. – mr_e_man Aug 7 '18 at 7:32
• @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 ? – JJacquelin Aug 7 '18 at 7:33

## 3 Answers

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

$\frac{dy}{dt}=10y-10t$

gives

$\frac{dy}{dt}-10y=-10t$

not

$\frac{dy}{dt}-10y=10t$

First step is to solve the homogeneous equation

$\frac{dy}{dt}-10y=0$

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

$\frac{du}{dt}-10u=-10t$

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 $b$.

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.

$$\frac{dy}{dt}=10y-10t$$ $$\frac{dy}{dt}-10y=-10t$$ 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 $$y(t)= t+\frac {1}{10}+Ke^{10t}$$