# Solving Differential Equation $y'=4e^{0.8t}-0.5y$

I'm having trouble with integrating this function. Could someone please add the steps to get to the answer? Initial condition $y(0) = 2$.

$y' = 4e^{0.8t}-0.5y$

$y= 4/1.3(e^{0.8t} - e^{-0.5t}) + 2e^{-0.5t}$

-
That answer is not right unless you are also given an initial condition $y(0)=2$. – Robert Israel Jul 29 '12 at 18:35
@Code-Guru, I got as far this with C=-3.\\ $y = 5e^{0.8t} - 0.5yt +C$ \\ $(1+.5t)y = 5e^{0.8t} + C$ \\ $y = (5e^{0.8t})/(1+0.5t) + C$ \\ $y=(5e^{.8t})/(1+.5t) + C$ Robert Israel, I've added the initial conditions – user994165 Jul 29 '12 at 19:04
Please edit your question to include your work. Also you should indicate if you have a specific question about something that you don't understand. – Code-Guru Jul 29 '12 at 19:17

Let $u(t) = e^{\int.5 dt} = e^{.5t}.$ This is our integrating factor. Multiplying the equation by $u(t)$ we get

$$e^{.5t}y'+.5e^{.5t}y = \frac{d}{dt}(e^{.5t}y) = 4e^{1.3t}$$

Integrating both sides yields

$$e^{.5t}y = \frac{4}{1.3}e^{1.3t} + C$$

So the general solution is

$$y = \frac{4}{1.3}e^{.8t} + Ce^{-.5t}$$

Now use the initial conditions to find $C$.

-

Outline: (i) Find the general solution of the homogeneous equation $y'+0.5y=0$ or equivalently $y'=-0.5y$. You probably know how to do this, it is a differential equation for exponential decay. Make sure that the expression you get includes all solutions of the homogeneous equation.

(ii) Find a particular solution of the inhomogeneous equation $y'+0.5y=4e^{0.8t}$. I suggest looking for a particular solution of the shape $Be^{0.8t}$, and substituting in $y'+0.5y=4e^{0.8t}$ to find the $B$ that works.

(iii) The general solution of your equation is then the general solution found in (i) plus the particular solution found in (ii).

(iv) Use your initial condition $y(0)=2$ to get the final answer.

-

As to how you get the integrating factor (nothing original here, but I'm just writing it explicitly):

If you want $y' + ay = f(x)$, use the fact that $(e^{ax}y)' = ae^{ax}y+e^{ax}y' = e^{ax}(y'+ay)$. Like magic, the left side of your equation appears.

So, multiplying your equation by $e^{ax}$ (which is, fortunately, always non-zero), $e^{ax}f(x) = e^{ax}(y'+ay) = (e^{ax}y)'$.

Integrating this, $e^{ax}y = \int e^{ax}f(x) \ dx$, or $y = e^{-ax}\int e^{ax}f(x) \ dx$.

This has the standard generalization to $y' + g(x)y = f(x)$.

-

We must realise that this kind of problem has homogeneous and particular solution.

Thus it helps to write the form of the equation given as below; $$(D+0.5)y = 4e^{0.8t}$$ D is the differential operator. Now this shows that (m+0.5) = 0, which gives m=-0.5. So the homogeneous solution is $$y_{h}= Ke^{0.8t}$$ Where K is a constant which need to solve by using the given initial condition.

Now we have to find the particular solution. It has to have some form of the derivatives of the forcing function i.e

$$4e^{0.8t}$$

lets assume that the particular solution has this form;

$$y_{p} = Ae^{0.8t}$$

putting this into the original equation;

$$D(Ae^{0.8t}) + 0.5(Ae^{0.8t}) = 4e^{0.8t}$$

so

$$0.8Ae^{0.8t} +0.5Ae^{0.8t} = 4e^{0.8t}$$ Which means

$$0.8A+0.5A = 4$$

so A = 4/1.3

Thus the particular solution is

$$y_{p} = 4/1.3 (e^{0.8t})$$

now putting everything together;

$$y = y_{h} + y_{p}$$

$$y = Ke^{-0.5t} + 4/1.3 e^{0.8t}$$

To find K, we have to use the initial condition given.

$$y(0) = 2 = Ke^{-0.5(0)} + 4/1.3 e^{0.8(0)}$$ $$2 = K(1) + 4/1.3 (1)$$ $$K = 2 - 4/1.3$$

Putting everything back into the solution;

$$y = 2e^{-0.5t} + (4/1.3) (e^{0.8t} -e^{-0.5t})$$

-