Register to reply 
How do you check your answer in a diff. eq. problem when you can't solve for Y? 
Share this thread: 
#1
Jan2713, 11:36 AM

P: 179

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


#2
Jan2713, 12:44 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 6,733

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



#3
Jan2713, 12:51 PM

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.



#4
Jan2713, 12:51 PM

Mentor
P: 21,397

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



#5
Jan2713, 01:04 PM

P: 179

It looks like I will need to provide example to explain the question more clearly.
Question: tan^{3}x*[itex]\frac{y'}{x}[/itex]=(y^{2}+4)sec^{2}x Answer: [itex]\frac{1}{2}[/itex]*sin^{2}x[itex]\frac{1}{4y}[/itex][itex]\frac{tan^1(y/2)}{8}[/itex] = 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? 


#6
Jan2713, 08:47 PM

Mentor
P: 21,397

For example, if you ended up with this equation... x + x^{2}sin(y) = 5 d/dx(x + x^{2}sin(y) ) = d(5)/dx => 1 + x^{2}cos(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. 


#7
Jan2713, 09:00 PM

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)*y^{2}*(y^{2}+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? 


#8
Jan2813, 12:36 AM

Mentor
P: 21,397

You differentiated (1/2)sin^{2}(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. 


#9
Jan2813, 08:52 AM

P: 179




#10
Jan2813, 09:05 AM

Mentor
P: 21,397

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. 


Register to reply 
Related Discussions  
Pendulum problem(check answer please?)  Introductory Physics Homework  1  
Buoyancy problem answer check. Thanks!  Introductory Physics Homework  1  
Forces on pin.. Did the problem but cant check answer  Engineering, Comp Sci, & Technology Homework  12  
Intial Value problem, Diff EQ, my steps look right but answer = wrong! wee!  Calculus & Beyond Homework  5 