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Math Help - Differential Equation Substitution

  1. #1
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    Differential Equation Substitution

    I'm trying to solve the problem (x+y+1)^2 dy/dx + (x+y+1)^2 + x^3 = 0. I've tried the substituting for x+y+1, x+y and (x+y+1)^2 but all to no avail. Could somebody please give a suggestion? This problem apparently just requires a change of variables and separation.
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  2. #2
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    Quote Originally Posted by StaryNight View Post
    I'm trying to solve the problem (x+y+1)^2 dy/dx + (x+y+1)^2 + x^3 = 0. I've tried the substituting for x+y+1, x+y and (x+y+1)^2 but all to no avail. Could somebody please give a suggestion? This problem apparently just requires a change of variables and separation.
    This equation is exact Note that

    \underbrace{(x+y+1)^2}_{M}dy+\underbrace{[(x+y+1)^2+x^3]}_{N}dx=0

    M_x=2(x+y+1)=N_y

    Can you finish from here?
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    This equation is exact Note that

    \underbrace{(x+y+1)^2}_{M}dy+\underbrace{[(x+y+1)^2+x^3]}_{N}dx=0

    M_x=2(x+y+1)=N_y

    Can you finish from here?
    I'm afraid i'm still confused...
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by StaryNight View Post
    I'm afraid i'm still confused...
    Here is a link
    Exact differential equation - Wikipedia, the free encyclopedia
    This should also be in your book.

    An equation is exact if there is a function f(x,y) such that

    \nabla f \cdot d\vec{r}=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}=0

    Where \vec{r}=x\vec{i}+y\vec{j} is the position vector.

    Since the mixed partials match we know such a function exists so

    \frac{\partial f}{\partial y}=(x+y+1)^2 \implies f(x,y)=\frac{1}{3}(x+y+1)^3+g(x)

    But we also know what the partial derivative with respect to x must be so we can solve for g(x)

    \frac{\partial f}{\partial x}=(x+y+1)^2+g'(x)=(x+y+1)^2+x^3 \implies g(x)=\frac{1}{4}x^4

    This gives \frac{1}{3}(x+y+1)^3+\frac{1}{4}x^4=c
    is an implicit solution to the equation.
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    Here is a link
    Exact differential equation - Wikipedia, the free encyclopedia
    This should also be in your book.

    An equation is exact if there is a function f(x,y) such that

    \nabla f \cdot d\vec{r}=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}=0

    Where \vec{r}=x\vec{i}+y\vec{j} is the position vector.

    Since the mixed partials match we know such a function exists so

    \frac{\partial f}{\partial y}=(x+y+1)^2 \implies f(x,y)=\frac{1}{3}(x+y+1)^3+g(x)

    But we also know what the partial derivative with respect to x must be so we can solve for g(x)

    \frac{\partial f}{\partial x}=(x+y+1)^2+g'(x)=(x+y+1)^2+x^3 \implies g(x)=\frac{1}{4}x^4

    This gives \frac{1}{3}(x+y+1)^3+\frac{1}{4}x^4=c
    is an implicit solution to the equation.
    So what exactly is the substituion you are meant to use?
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  6. #6
    A Plied Mathematician
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    You said in your OP that this problem "apparently just requires a change of variables and separation." Does that mean you are required to solve it that way? Or could you solve it using the exact differential method that TheEmptySet has outlined for you? Because the exact differential method that TheEmptySet has outlined for you is not a substitution method, nor is it a separation of variables idea.
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    You said in your OP that this problem "apparently just requires a change of variables and separation." Does that mean you are required to solve it that way? Or could you solve it using the exact differential method that TheEmptySet has outlined for you? Because the exact differential method that TheEmptySet has outlined for you is not a substitution method, nor is it a separation of variables idea.
    The question reads "Solve by a change of variables and separation". I'm in my first year of university and haven't covered the method TheEmptySet proposed.
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  8. #8
    A Plied Mathematician
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    Hmm. Why don't you show your work for the substitution u = x + y + 1?
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  9. #9
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    yes, the same, it's pretty straightforward, and it yields a separable equation.
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  10. #10
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    Quote Originally Posted by Ackbeet View Post
    Hmm. Why don't you show your work for the substitution u = x + y + 1?
    Having had another go at the problem with this substitution I've managed to get a solution.

    Thanks for your help.
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  11. #11
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    You're very welcome. Have a good one!
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