# How do you check your answer in a diff. eq. problem when you can't solve for Y?

• Jeff12341234
In summary, you need to differentiate both sides of the equation to check if it's correct. If it's not, you'll need to use implicit differentiation to solve for y.
Jeff12341234
How do you check your answer in a diff. eq. problem when you can't solve for Y?

I have a Ti-nSpire CAS. Is there a graphical or other way to check that you have the right answer when solving a D.E. when you CAN'T solve for y?

Your comment is not clear. If you can't solve for Y, how can you check that Y is the right answer?

I'm new to D.E but from what I've learned so far, there are some that can't be 'explicitly' solved for y. The answer contains y (the dependent var) in several different places at once like within trig functions while also being in something simple like a fraction. That's the scenario I'm referring to.

Jeff12341234 said:
How do you check your answer in a diff. eq. problem when you can't solve for Y?
It sounds like your solution defines y implicitly as a function of x (or t, or whatever the independent variable is). To check your solution, differentiate both sides of your equation implicitly, and solve for dy/dt, then substitute into your differential equation.
Jeff12341234 said:
I have a Ti-nSpire CAS. Is there a graphical or other way to check that you have the right answer when solving a D.E. when you CAN'T solve for y?

It looks like I will need to provide example to explain the question more clearly.
Question: tan3x*$\frac{y'}{x}$=(y2+4)sec2x

Answer: $\frac{1}{2}$*sin2x-$\frac{1}{4y}$-$\frac{tan^-1(y/2)}{8}$ = c

As far as I know, I need to be able to solve for Y to be able to plug it back into check it. So what do you do when you can't solve for Y?

Jeff12341234 said:
It looks like I will need to provide example to explain the question more clearly.
Question: tan3x*$\frac{y'}{x}$=(y2+4)sec2x

Answer: $$\frac{1}{2}*sin^2(x)-\frac{1}{4y}-\frac{tan^{-1}(y/2)}{8} = c$$

As far as I know, I need to be able to solve for Y to be able to plug it back into check it.
No, and my answer is the same as before. Differentiate both sides of the equation above implicitly, and solve algebraically for dy/dx.

For example, if you ended up with this equation...

x + x2sin(y) = 5

d/dx(x + x2sin(y) ) = d(5)/dx
=> 1 + x2cos(y)*dy/dx + 2x sin(y) = 0

Now group all terms that involve dy/dx on one side, and all other terms on the right.
Solve for dy/dx.
Jeff12341234 said:
So what do you do when you can't solve for Y?
See above.

I don't completely follow. I differentiated my answer above in terms of x and got sin(x)*cos(x)=0. If I differentiate it implicitly with y as dependent I get -sin(x)*cos(x)*y2*(y2+4) What do I do with that?

Is there a link to a worked example where they start from an answer similar to what I got, then end up with Y=something, then plug that back into the original D.E. to check it?

Last edited:
Jeff12341234 said:
I don't completely follow. I differentiated my answer above in terms of x and got sin(x)*cos(x)=0.
That's incorrect because you didn't differentiate the 1/(4y) and (1/8)tan-1(y/2) terms. When you differentiate implicitly, you differentiate everything with respect to the same variable, x in this case.

You differentiated (1/2)sin2(x) correctly, but to differentiate the terms in y you have to use the chain rule. So in the above, to differentiate 1/(4y) or (1/4)y-1, with respect to x, you have

d/dx[(1/4)y-1) = 1/4 * (-1) y-2 * dy/dx. The last factor comes from the chain rule. The same sort of thing needs to happen with the tan-1 term.

When you were learning about derivatives, there should have been a section on implicit differentiation.
Jeff12341234 said:
If I differentiate it implicitly with y as dependent I get -sin(x)*cos(x)*y2*(y2+4) What do I do with that?
I have no idea how you got this.
Jeff12341234 said:
Is there a link to a worked example where they start from an answer similar to what I got, then end up with Y=something, then plug that back into the original D.E. to check it?
The approach you want is applicable only if you have very simple equations in which you can actually solve for y. Many times this is not possible, but you can follow the approach I have suggested throughout this thread - use implicit differentiation. If you don' don't know it, or don't remember learning it, or have forgotten it, look it up.

Mark44 said:
The approach you want is applicable only if you have very simple equations in which you can actually solve for y. Many times this is not possible, ...
Ah, see, that is the problem. I need a way to check my answer even if y isn't solvable. I need a way to know for sure if my general form answer is right.

Jeff12341234 said:
It looks like I will need to provide example to explain the question more clearly.
Question: tan3x*$\frac{y'}{x}$=(y2+4)sec2x (1)

Answer: $\frac{1}{2}$*sin2x-$\frac{1}{4y}$-$\frac{tan^-1(y/2)}{8}$ = c (2)

As far as I know, I need to be able to solve for Y to be able to plug it back into check it. So what do you do when you can't solve for Y?

Jeff12341234 said:
Ah, see, that is the problem. I need a way to check my answer even if y isn't solvable. I need a way to know for sure if my general form answer is right.

Using the problem you posted as an example, here's how you check it.
Solve algebraically for dy/dx.

Substitute for dy/dx in the original differential equation - (1). If your solution is correct, you should get an equation that is identically true.

You DON'T need to solve for y in the solution.

## 1. How do I know if my answer is correct if I can't solve for Y in a differential equation problem?

There are a few ways to check your answer in a differential equation problem. One method is to plug your answer into the original equation and see if it satisfies the equation. Another approach is to graph both the original equation and your answer to see if they match. Additionally, you can use numerical methods such as Euler's method or Runge-Kutta methods to approximate the solution and compare it to your answer.

## 2. Can I use software or calculators to check my answer in a diff. eq. problem?

Yes, you can use software or calculators to check your answer in a differential equation problem. There are many software programs specifically designed for solving differential equations, such as MATLAB or WolframAlpha. These programs can also graph the solutions and compare them to your answer. However, it is important to also understand the concepts and techniques used to solve differential equations manually.

## 3. What should I do if my answer doesn't match the solutions provided in the textbook or by the teacher?

If your answer doesn't match the solutions provided, there are a few steps you can take to troubleshoot. First, double check your work and make sure you didn't make any mistakes in your calculations. Then, try using a different method to solve the differential equation to see if you get the same answer. You can also ask your teacher or classmates for help or seek out additional resources, such as online tutorials or textbooks, to better understand the problem and solution.

## 4. Is it possible to have multiple correct answers for a differential equation problem?

Yes, it is possible to have multiple correct answers for a differential equation problem. This is because differential equations can have families of solutions, meaning there are multiple possible functions that can satisfy the equation. In some cases, the general solution may also include arbitrary constants that can result in a variety of solutions. It is important to understand the concept of general solutions and initial conditions in order to find a specific solution.

## 5. What should I do if I am still unsure about my answer after checking it?

If you are still unsure about your answer after checking it, it is best to seek out additional help from your teacher or a classmate. You can also consult online resources or textbooks for further explanations and examples. It is important to fully understand the concepts and techniques used in solving differential equations in order to confidently check your answers and solve future problems.

Replies
4
Views
1K
Replies
6
Views
1K
Replies
3
Views
3K
Replies
2
Views
3K
Replies
3
Views
2K
Replies
16
Views
3K
Replies
6
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
65
Views
3K