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

#### bigwave

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

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

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

#### romsek

MHF Helper
it doesn't seem to be a solution.

Are there any typos?

1 person

its number 13

#### MarkFL

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 $$\displaystyle \cos(t)$$.

2 people

#### bigwave

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
$$y''=(y')^2$$

#### MarkFL

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:

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

3 people

#### bigwave

whoa....

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

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#### MarkFL

whoa....

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

1 person