Hey there! Sign in to join this conversationNew here? Join for free

General solution of first order ODE stuck on integrating Watch

    • Thread Starter
    Offline

    6
    ReputationRep:
    I'm stuck on the following differential equation question, I have to solve it and find a general solution in explicit form:
    \displaystyle\frac{dy}{dx} = \sqrt{(1-y^2)} \times sin(x)

    My attempt is as follows, using the variable separable method:
    \displaystyle\frac{1}{(1-y^2)^\frac{1}{2}} \frac{dy}{dx} = sin(x)

    \displaystyle\int \frac{1}{(1-y^2)^\frac{1}{2}} dy = \int sin(x) dx

    Here is where I got stuck integrating the LHS. Am I correct in integrating by substitution with y = sinx?

    I tried it but the answer I got looked wrong, I got upto (for LHS):
    \displaystyle\int sec^2(x) dy

    I'm not sure whether my integration by trig substitution is the problem or if this type of ODE doesn't work for the variable separation method, any insights would be appreciated.
    Offline

    3
    ReputationRep:
    For the substitution, you will need to introduce another variable, so y = \sin u is OK. Then recall that \frac{d}{du}\tan u = \sec^2 u.
    Online

    11
    ReputationRep:
    (Original post by Sayless)
    I'm stuck on the following differential equation question, I have to solve it and find a general solution in explicit form:
    \displaystyle\frac{dy}{dx} = \sqrt{(1-y^2)} \times sin(x)

    My attempt is as follows, using the variable separable method:
    \displaystyle\frac{1}{(1-y^2)^\frac{1}{2}} \frac{dy}{dx} = sin(x)

    \displaystyle\int \frac{1}{(1-y^2)^\frac{1}{2}} dy = \int sin(x) dx

    Here is where I got stuck integrating the LHS. Am I correct in integrating by substitution with y = sinx?

    I tried it but the answer I got looked wrong, I got upto (for LHS):
    \displaystyle\int sec^2(x) dy

    I'm not sure whether my integration by trig substitution is the problem or if this type of ODE doesn't work for the variable separation method, any insights would be appreciated.

    That is the correct substitution, however it seems like you forgot to square root the denominator after the sub

    Also remember that if y=sinx then dy/dx = cosx and so dy=cosx dx meaning the numerator and denominator should end up cancelling out

    Or just use the formula book since it's a standard integral
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by A Slice of Pi)
    For the substitution, you will need to introduce another variable, so y = \sin u is OK. Then recall that \frac{d}{du}\tan u = \sec^2 u.
    with \displaystyle\int sec^2(x) dy
    I have to integrate with respect to y, and I have an x value so I imagined it as a constant and having a 1 infront of it so turning that into a y with the constant following, I'm not sure what to do here.
    I can see what you mean and I would do that if the integral was with respect to x, do you end up with my answer when you try the substitution method?
    • Thread Starter
    Offline

    6
    ReputationRep:
    Name:  Photo 04-02-2017 18 51 07.jpg
Views: 32
Size:  485.3 KB
    I've attached my working here, did I substitute it right?
    Offline

    22
    ReputationRep:
    (Original post by Sayless)
    I've attached my working here, did I substitute it right?
    No, this is what happens when you get too used to what "x" is meant to mean.

    You want to substitute for dy in terms of dx, because your original integral is with respect to dy and you want to change it to dx. So dy = cos x dx.

    Then \int \frac{1}{\cos x} \, \mathrm{d}y = \int \frac{1}{\cos x} \cdot \cos x \, \mathrm{d}x = \int 1 \, \mathrm{d}x

    If I were you, I'd just avoid using x as a substitution variable and stick with u or something...
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by Zacken)
    No, this is what happens when you get too used to what "x" is meant to mean.

    You want to substitute for dy in terms of dx, because your original integral is with respect to dy and you want to change it to dx. So dy = cos x dx.

    Then \int \frac{1}{\cos x} \, \mathrm{d}y = \int \frac{1}{\cos x} \cdot \cos x \, \mathrm{d}x = \int 1 \, \mathrm{d}x

    If I were you, I'd just avoid using x as a substitution variable and stick with u or something...
    Ah, I see where I went wrong now thanks for pointing it out. You're right I'm too used to subbing the dy into the dx so I did it out of habit without thinking about it, I'll begin using u instead
    Online

    14
    ReputationRep:
    (Original post by Sayless)
    Name:  Photo 04-02-2017 18 51 07.jpg
Views: 32
Size:  485.3 KB
    I've attached my working here, did I substitute it right?
    use the fact that \int\frac{1}{\sqrt{1-y^{2}}}\text{d}y=\text{arcsin}(y  )+c
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by Cryptokyo)
    use the fact that \int\frac{1}{\sqrt{1-y^{2}}}\text{d}y=\text{arcsin}(y  )+c
    Name:  Photo 04-02-2017 19 15 57.jpg
Views: 28
Size:  491.1 KB
    I'm trying that now and I got the general solution in explicit form but I'm having trouble verifying it, I've attached my working.
    I need the sqrt(1-y^2) to be sin(C-cosx) to verify it but I can only get it in terms of cos, am I doing it right?

    if my working is unclear i can rewrite it clearly including full question, it's all scattered atm, the question is in the first post
    Online

    14
    ReputationRep:
    (Original post by Sayless)
    Name:  Photo 04-02-2017 19 15 57.jpg
Views: 28
Size:  491.1 KB
    I'm trying that now and I got the general solution in explicit form but I'm having trouble verifying it, I've attached my working.
    I need the sqrt(1-y^2) to be sin(C-cosx) to verify it but I can only get it in terms of cos, am I doing it right?

    if my working is unclear i can rewrite it clearly including full question, it's all scattered atm, the question is in the first post
    Sorry for the late reply.

    You differentiation is incorrect. The derivative of \sin(C-\cos(x)) is \sin x \times \cos(C-\cos(x)) and not \sin x \times \sin(C-\cos (x)) It should work then.
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by Cryptokyo)
    Sorry for the late reply.

    You differentiation is incorrect. The derivative of \sin(C-\cos(x)) is \sin x \times \cos(C-\cos(x)) and not \sin x \times \sin(C-\cos (x)) It should work then.
    Ah I've got it now thank you,
    Offline

    3
    ReputationRep:
    1/a ln(ax+b) is the solution for integration a fraction. When a is 1 and the numerator is one then the integral is just ln|f(x)|
    Offline

    17
    ReputationRep:
    (Original post by Anfanny)
    1/a ln(ax+b) is the solution for integration a fraction. When a is 1 and the numerator is one then the integral is just ln|f(x)|
    It's the general solution for integrating 1/(ax+b).

    The integral here is not of that form.
    Offline

    22
    ReputationRep:
    (Original post by Cryptokyo)
    use the fact that \int\frac{1}{\sqrt{1-y^{2}}}\text{d}y=\text{arcsin}(y  )+c
    And why's that true...?
    Online

    14
    ReputationRep:
    Sayless You may be interested in this...

    (Original post by Zacken)
    And why's that true...?
    Let y=\text{arcsin}(x)
    \Rightarrow x=\sin y
    \frac{dx}{dy}=\cos y
    Using \frac{dy}{dx}\frac{dx}{dy}=1
    \frac{dy}{dx}= \frac{1}{\cos y}= \frac{1}{\sqrt{1-\sin^{2}y}}= \frac{1}{\sqrt{1-x^{2}}}
    Offline

    17
    ReputationRep:
    (Original post by Cryptokyo)
    Sayless You may be interested in this...



    Let y=\text{arcsin}(x)
    \Rightarrow x=\sin y
    \frac{dx}{dy}=\cos y
    Using \frac{dy}{dx}\frac{dx}{dy}=1
    \frac{dy}{dx}= \frac{1}{\cos y}= \frac{1}{\sqrt{1-\sin^{2}y}}= \frac{1}{\sqrt{1-x^{2}}}
    Technically, this is correct. But (I'm sure) the point Zacken is trying to make here is that you should be able to find this result by substitution.

    To be clear, instead of using substitution, you can do this by recognition (which is kind of what you've done, otherwise you'd have no reason to "start" with arcsin x), or by looking it up in the formula book, and these are all fine valid methods.

    But if someone tries to do this by substitution, and they can't, then they need to brush up on your substitution methods; someone telling them "it's easy to verify that arcsin x works" is not doing them a service in the long term.
    Offline

    3
    ReputationRep:
    (Original post by Sayless)
    with \displaystyle\int sec^2(x) dy
    I have to integrate with respect to y, and I have an x value so I imagined it as a constant and having a 1 infront of it so turning that into a y with the constant following, I'm not sure what to do here.
    I can see what you mean and I would do that if the integral was with respect to x, do you end up with my answer when you try the substitution method?
    You have made the substitution  y = \sin x and your intuition is correct. However, you will need to use another variable not aready present, say u, instead of x. Remember the goal here. You're solving the ODE to find y as a function of x. When you write  y = \sin x you're actually claiming that this is the solution of the ODE!
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by A Slice of Pi)
    You have made the substitution  y = \sin x and your intuition is correct. However, you will need to use another variable not aready present, say u, instead of x. Remember the goal here. You're solving the ODE to find y as a function of x. When you write  y = \sin x you're actually claiming that this is the solution of the ODE!
    Ah I used x because the RHS was in terms of x so I thought it would be easier, it still works with x but for better presentation I should use u instead?
    Offline

    3
    ReputationRep:
    (Original post by Sayless)
    Ah I used x because the RHS was in terms of x so I thought it would be easier, it still works with x but for better presentation I should use u instead?
    No, we use u instead of x because x is already present on the RHS. Never use x for substitutions if it is already present somewhere else!
    Offline

    22
    ReputationRep:
    (Original post by DFranklin)
    Technically, this is correct. But (I'm sure) the point Zacken is trying to make here is that you should be able to find this result by substitution.
    PRSOM
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.