How to Solve a Tough Differential Equation on a Calc I Exam?

[Dethrone]

Solve:
$$\sin\left({x}\right)y''+(y'^2-\sin^2\left({x}\right))^{1/2}y'^2-\cos\left({x}\right)y'=0$$

Initial conditions:
$y(0)=0$
$y'(0)=1$

Keep in mind that this question was on a Calc I exam worth 5 marks, so please nothing crazy like reduction of order or anything...:D

 

[Opalg]

Solve:
$$\sin\left({x}\right)y''+(y'^2-\sin^2\left({x}\right))^{1/2}y'^2-\cos\left({x}\right)y'=0$$

Initial conditions:
$y(0)=0$
$y'(0)=1$

Keep in mind that this question was on a Calc I exam worth 5 marks, so please nothing crazy like reduction of order or anything...:D

Given that this question was on a Calc I exam worth 5 marks, I can only assume that you are meant to guess that the solution is $y=x$. You can easily verify that this solution works for $x$ in the interval $\bigl[-\frac\pi2, \frac\pi2\bigr]$ (but not outside that interval, because then $(1-\sin^2x)^{1/2}$ would no longer be equal to $\cos x$).

Now I'm wondering what sort of instructor would set such a question on a Calc I exam?

 

[MarkFL]

Given that this question was on a Calc I exam worth 5 marks, I can only assume that you are meant to guess that the solution is $y=x$. You can easily verify that this solution works for $x$ in the interval $\bigl[-\frac\pi2, \frac\pi2\bigr]$ (but not outside that interval, because then $(1-\sin^2x)^{1/2}$ would no longer be equal to $\cos x$).

Now I'm wondering what sort of instructor would set such a question on a Calc I exam?

The OP stated earlier in our live chat that:

...there's a chance I might not get back to you until after exams...

So, I will take it upon myself to state he also had this to say:

They say the solution is $y=x$, but no marks for simply guessing.

My prof is very evil...hehe

He told us that when he was a student, he swore to himself that when he became a prof, he would write on the chalkboard faster than any of his profs, and make tests harder than the ones he's had.

But he's a really cool guy...(Cool)

 

[Ackbach]

So your IVP is
\begin{align*}
\sin\left({x}\right)y''+(y'^2-\sin^2\left({x}\right))^{1/2}y'^2-\cos\left({x}\right)y'&=0 \\
y(0)&=0 \\
y'(0)&=1.
\end{align*}

The first thing I would notice is that there is no $y$. So you could let $z=y'$, and obtain the first-order equation
$$\sin\left({x}\right)z'+(z^2-\sin^2\left({x}\right))^{1/2}z^2-\cos\left({x}\right)z=0,$$
with IC $z(0)=1$.

At this point, you could observe about domains. If you're looking for a solution on an interval that contains $\pi/4\pm 2 k\pi,$ then $z$ is forced to be identically $1$, or else you would dive into the complex world, what with the $\sqrt{z^2-\sin^2(x)}$ and all.

 

[Dethrone]

Hi Ackbach!

So the only way to solve the resulting first-order DE is by inspection of domain?

$$\sin\left({x}\right)z'+(z^2-\sin^2\left({x}\right))^{1/2}z^2-\cos\left({x}\right)z=0$$

Also, should not the $\sin\left({x}\right)z'$ in the first term be $\sin\left({x}\right)z' \cdot z$?

Since \(\displaystyle z=y'\), so \(\displaystyle y''=\d{z}{x}=\d{z}{y}\cdot \d{y}{x}=z' \cdot y'=z'z\).

EDIT: You are correct: it should be $\sin\left({x}\right)z'$, as we are working with respect to $x$.