# Solving First Order Differential Equation

1. Mar 4, 2014

### jellicorse

1. The problem statement, all variables and given/known data

I have been trying to solve this equation but keep coming to the same solution, which according to my book is not the correct one. Is anyone able to point out what I am doing wrong?

$$\frac{dy}{dt}-\frac{1}{2}y=2cos(t)$$

3. The attempt at a solution

To solve, use the integrating factor $$e^{\int\frac{1}{2}dt}$$; Integrating factor=$$e^{-\frac{t}{2}}$$

$$ye^{-\frac{t}{2}} = 2\int e^{-\frac{t}{2}}cos(t)dt$$

Integrating the RHS by parts:

$$= 2\left[e^{-\frac{t}{2}}sin(t)+\frac{1}{2}\int sin(t)e^{-\frac{t}{2}}dt\right]$$

$$= 2\left[e^{-\frac{t}{2}}sin(t)+\frac{1}{2}\left[e^{-\frac{t}{2}}\cdot(-cos(t)-\frac{1}{2}\int cos(t)e^{-\frac{t}{2}}dt\right]\right]$$

And using a reduction formula:

$$= 2\left[e^{-\frac{t}{2}}sin(t)-\frac{e^{-\frac{t}{2}}cos(t)}{2}-\frac{1}{4}I\right]$$

$$I =2e^{-\frac{t}{2}}sin(t)-e^{-\frac{t}{2}}cos(t)-\frac{1}{2}I$$

$$\frac{3}{2}I=2e^{-\frac{t}{2}}sin(t)-e^{-\frac{t}{2}}cos(t)$$

$$I = \frac{2}{3}(2e^{-\frac{t}{2}}sin(t)-e^{-\frac{t}{2}}cos(t))$$

$$ye^{-\frac{t}{2}}=\frac{4e^{-\frac{t}{2}}sin(t)-e^{-\frac{t}{2}}cos(t)}{3}$$

$$y = \frac{4 sin(t)-cos(t)}{3}+ce^{\frac{t}{2}}$$

After all this, the book gives a solution of $$y=\frac{4}{5}(2sin(t)-cos(t))+ce^{\frac{t}{2}}$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 4, 2014

### tiny-tim

hi jellicorse! welcome to pf!

good method, but i think you've used ye-t/2 = 2I in one place and = I in another place

3. Mar 4, 2014

### jellicorse

Ah, thanks tiny-tim... I will look into that!