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

In summary, the given equation is a first-order differential equation that can be simplified by letting $z=y'$ and then solving for $z$ using the initial conditions. However, the solution $y=x$ only works for certain intervals and guessing is not an acceptable method for solving this problem on an exam. The instructor's intention may have been to test the students' understanding of domains and solutions to differential equations.
  • #1
Dethrone
717
0
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
 
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  • #2
Rido12 said:
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?
 
  • #3
Opalg said:
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:

Rido12 said:
...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:

Rido12 said:
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)
 
  • #4
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.
 
  • #5
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 $x$.
 
Last edited:

1. What is a tough differential equation?

A tough differential equation is a mathematical equation that involves derivatives of an unknown function. It is considered "tough" because it is generally difficult to solve analytically, meaning there is no exact formula for the solution.

2. Why are differential equations important in science?

Differential equations are important in science because they are used to model and describe many natural phenomena, such as the growth of populations, the movement of objects, and the change in temperature over time. They allow scientists to make predictions and understand complex systems.

3. What makes solving tough differential equations challenging?

Solving tough differential equations can be challenging because they often involve complex mathematical operations, may not have an analytical solution, and can have multiple variables and parameters that need to be considered. Additionally, the initial conditions and boundary conditions may not be known, making it difficult to find the exact solution.

4. What methods are used to solve tough differential equations?

There are various methods used to solve tough differential equations, including numerical methods such as Euler's method and Runge-Kutta methods, as well as analytical methods such as separation of variables and Laplace transforms. The choice of method depends on the specific equation and the desired level of accuracy.

5. How are differential equations used in real-world applications?

Differential equations are used in a wide range of real-world applications, including engineering, physics, biology, economics, and more. They are used to model and predict the behavior of systems in these fields, such as the flight of airplanes, the spread of diseases, and the flow of electricity. Solving differential equations allows scientists and engineers to make accurate predictions and design efficient systems.

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