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

by Jeff12341234
Tags: check, diff, solve
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 P: 179 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?
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,546 Your comment is not clear. If you can't solve for Y, how can you check that Y is the right answer?
 P: 179 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.
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How do you check your answer in a diff. eq. problem when you can't solve for Y?

 Quote by Jeff12341234 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.
 Quote by Jeff12341234 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?
 P: 179 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 in to check it. So what do you do when you can't solve for Y?
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 Quote by Jeff12341234 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 in to 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.
 Quote by Jeff12341234 So what do you do when you can't solve for Y?
See above.
 P: 179 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?
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 Quote by Jeff12341234 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.
 Quote by Jeff12341234 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.
 Quote by Jeff12341234 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.
P: 179
 Quote by Mark44 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.
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 Quote by Jeff12341234 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 in to check it. So what do you do when you can't solve for Y?
 Quote by Jeff12341234 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.
Start with your solution equation - (2), and differentiate it implicitly.
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.

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