# Thread: 1.3.13 Verify that the given function is a solution of the differential equation

1. ## 1.3.13 Verify that the given function is a solution of the differential equation

$\textsf{Verify the following that the given function is a solution of the differential equation}$

$y^{\prime\prime}+y=\sec t \quad 0<t<\pi/2 \quad y=(\cos{t})\ln{t}+t\sin{t}$

ok presume this is a plug in thing
but the y'' is a little daunting

2. ## Re: 1.3.13 Verify that the given function is a solution of the differential equation

it doesn't seem to be a solution.

Are there any typos?

3. ## Re: 1.3.13 Verify that the given function is a solution of the differential equation

its number 13

4. ## Re: 1.3.13 Verify that the given function is a solution of the differential equation

The given solution is:

$\displaystyle y=\cos(t)\ln(\cos(t))+t\sin(t)$

You initially stated the argument for the log function was $\displaystyle t$ instead of $\displaystyle \cos(t)$.

5. ## Re: 1.3.13 Verify that the given function is a solution of the differential equation

ok so if
$$\displaystyle y=\cos(t)\ln(\cos(t))+t\sin(t)$$
then
$$\displaystyle y'= t \cos(t) - \sin(t) \log(\cos(t))$$
and
$$y''=(y')^2$$

6. ## Re: 1.3.13 Verify that the given function is a solution of the differential equation

I get:

$\displaystyle y=\cos(t)\ln(\cos(t))+t\sin(t)$

$\displaystyle y'=-\sin(t)-\sin(t)\ln(\cos(t))+t\cos(t)+\sin(t)=-\sin(t)\ln(\cos(t))+t\cos(t)$

$\displaystyle y''=\sin^2(t)\sec(t)-\cos(t)\ln(\cos(t))-t\sin(t)+\cos(t)=\cos(t)+\sin^2(t)\sec(t)-y=(\cos^2(t)+\sin^2(t))\sec(t)-y=\sec(t)-y$

And thus:

$\displaystyle y''+y=\sec(t)$

7. ## Re: 1.3.13 Verify that the given function is a solution of the differential equation

whoa....

not sure how you got $y''$ is that from $W\vert A$

8. ## Re: 1.3.13 Verify that the given function is a solution of the differential equation

Originally Posted by bigwave
whoa....

not sure how you got $y''$ is that from $W\vert A$
No, I just did it by hand with some mental simplifications.