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Math Help - Differential Equations

  1. #1
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    Differential Equations

    1.Obtain the solution to the differential equation

    dy/dx=(1+y^2)/(1+x^2)



    2. The first order differential equation

    dy/dx-xy=x

    has the solution y= ?

    These are both separable.
    1) \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}

    2) \frac{dy}{dx} - xy = x \rightarrow \frac{dy}{dx} = x(1+y) \rightarrow \frac{dy}{1 + y} = x dx


    Someone kindly finish up for me..I kind of confused at this stage.
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  2. #2
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    Quote Originally Posted by KayPee View Post
    1.Obtain the solution to the differential equation

    dy/dx=(1+y^2)/(1+x^2)


    2. The first order differential equation

    dy/dx-xy=x

    has the solution y= ?

    These are both separable.
    1) \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}

    2) \frac{dy}{dx} - xy = x \rightarrow \frac{dy}{dx} = x(1+y) \rightarrow \frac{dy}{1 + y} = x dx


    Someone kindly finish up for me..I kind of confused at this stage.

    1)  \frac{dy}{1+y^2} = \frac{dx}{1+x^2}

     \int \frac{dx}{x^2+a^2} = \frac{1}{a} \arctan{\frac{x}{a}} +C

    2)  \int \frac{dy}{1+y} = \ln{|1+y|}+C
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  3. #3
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    Can I have the final answer please!

    Quote Originally Posted by Mush View Post
    1)  \frac{dy}{1+y^2} = \frac{dx}{1+x^2}

     \int \frac{dx}{x^2+a^2} = \frac{1}{a} \arctan{\frac{x}{a}} +C

    2)  \int \frac{dy}{1+y} = \ln{|1+y|}+C

    I would appreciate it if I could have the final answer.

    I'm not getting it besides.I'm not used to the Latex
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  4. #4
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    Quote Originally Posted by KayPee View Post
    1.Obtain the solution to the differential equation

    dy/dx=(1+y^2)/(1+x^2)



    2. The first order differential equation

    dy/dx-xy=x

    has the solution y= ?

    These are both separable.
    1) \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}

    2) \frac{dy}{dx} - xy = x \rightarrow \frac{dy}{dx} = x(1+y) \rightarrow \frac{dy}{1 + y} = x dx


    Someone kindly finish up for me..I kind of confused at this stage.
    You've been given the solution already, but I'll go step by step...

    1. \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}

    \frac{1}{1 + y^2}\,\frac{dy}{dx} = \frac{1}{1 + x^2}

    \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{1 + x^2}\,dx}

    \int{\frac{1}{1 + y^2}\,dy}= \int{\frac{1}{1 + x^2}\,dx}

    \arctan{y} + C_1 = \arctan{x} + C_2

    \arctan{y} = \arctan{x} + C.


    2. \frac{dy}{dx} - xy = x

    \frac{dy}{dx} = xy + x

    \frac{dy}{dx} = x(y + 1)

    \frac{1}{y + 1}\,\frac{dy}{dx} = x

    \int{\frac{1}{y + 1}\,\frac{dy}{dx}\,dx} = \int{x\,dx}

    \int{\frac{1}{y + 1}\,dy} = \int{x\,dx}

    \ln{|y + 1|} + C_1 = \frac{1}{2}x^2 + C_2

    \ln{|y + 1|} = \frac{1}{2}x^2 + C

    |y + 1| = e^{\frac{1}{2}x^2 + C}

    |y + 1| = e^C e^{\frac{1}{2}x^2}

    y + 1 = \pm e^C e^{\frac{1}{2}x^2}

     y + 1 = Ae^{\frac{1}{2}x^2}

     y = Ae^{\frac{1}{2}x^2} - 1.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    You've been given the solution already, but I'll go step by step...

    1. \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}

    \frac{1}{1 + y^2}\,\frac{dy}{dx} = \frac{1}{1 + x^2}

    \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{1 + x^2}\,dx}

    \int{\frac{1}{1 + y^2}\,dy}= \int{\frac{1}{1 + x^2}\,dx}

    \arctan{y} + C_1 = \arctan{x} + C_2

    \arctan{y} = \arctan{x} + C.


    2. \frac{dy}{dx} - xy = x

    \frac{dy}{dx} = xy + x

    \frac{dy}{dx} = x(y + 1)

    \frac{1}{y + 1}\,\frac{dy}{dx} = x

    \int{\frac{1}{y + 1}\,\frac{dy}{dx}\,dx} = \int{x\,dx}

    \int{\frac{1}{y + 1}\,dy} = \int{x\,dx}

    \ln{|y + 1|} + C_1 = \frac{1}{2}x^2 + C_2

    \ln{|y + 1|} = \frac{1}{2}x^2 + C

    |y + 1| = e^{\frac{1}{2}x^2 + C}

    |y + 1| = e^C e^{\frac{1}{2}x^2}

    y + 1 = \pm e^C e^{\frac{1}{2}x^2}

     y + 1 = Ae^{\frac{1}{2}x^2}

     y = Ae^{\frac{1}{2}x^2} - 1.
    You missed a step (lol! Nice job with the latex, by the way).

    To the OP:

    1. Now, the least you can do in return is to confirm these solutions by substitution.

    2. The solution given to you for the first question can be expressed in a simpler form .....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    You missed a step (lol! Nice job with the latex, by the way).

    To the OP:

    1. Now, the least you can do in return is to confirm these solutions by substitution.

    2. The solution given to you for the first question can be expressed in a simpler form .....
    I can't be bothered trying to get the first one to read y = \dots because otherwise I'll be trying to find the most elegant way for it to read hahahaha.
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