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Math Help - Differential Equations (3 questions)

  1. #1
    Junior Member Lonehwolf's Avatar
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    Differential Equations (3 questions)

    1 Obtain the General Solution of the differential equations:
    (i) x\frac{dy}{dx} + y = xe^{3x}

    (ii) \frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}

    2 (a) Find the general solution of the differential equation
    x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x

    (b) Find the solution of

    \frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}

    given that at x=0,y=2 and \frac{dy}{dx}=4

    3 (a) Find the following two integrals:
    \int\frac{x}{1-x^2}dx and \int\frac{x}{1-x^2}dx

    Find y in terms of x, givene that
    \frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}

    and that y =1 when x = 0.

    (b) Solve the differential equation
    \frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0

    given that y=1 and \frac{dy}{dx}=3 when x=0.
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  2. #2
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    Quote Originally Posted by Lonehwolf View Post
    1 Obtain the General Solution of the differential equations:
    (i) x\frac{dy}{dx} + y = xe^{3x}

    (ii) \frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}

    2 (a) Find the general solution of the differential equation
    x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x

    (b) Find the solution of

    \frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}

    given that at x=0,y=2 and \frac{dy}{dx}=4


    3 (a) Find the following two integrals:
    \int\frac{x}{1-x^2}dx and \int\frac{x}{1-x^2}dx

    Find y in terms of x, givene that
    \frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}

    and that y =1 when x = 0.

    (b) Solve the differential equation
    \frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0

    given that y=1 and \frac{dy}{dx}=3 when x=0.
    1. Notice that the left hand side is the product-rule expansion of

    \frac{d}{dx}(xy)

    So \frac{d}{dx}(xy) = xe^{3x}

    xy = \int{xe^{3x}\,dx}.

    You can evaluate the RHS using integration by parts.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Lonehwolf View Post
    1 Obtain the General Solution of the differential equations:
    (i) x\frac{dy}{dx} + y = xe^{3x}
    Divide through by the coefficient of \frac{\,dy}{\,dx} to get \frac{\,dy}{\,dx}+\frac{1}{x}y=e^{3x}

    Apply the technique of the integrating factor here. Can you continue?

    (ii) \frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}
    Do you mean \frac{\,d^2y}{\,dx^2}{\color{red}-}5\frac{\,dy}{\,dx}+6y=10e^{-2x} or \frac{\,d^2y}{\,dx^2}{\color{red}+}5\frac{\,dy}{\,  dx}+6y=10e^{-2x}??

    2 (a) Find the general solution of the differential equation
    x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x
    Divide through by the coefficient of \frac{\,dy}{\,dx} to get \frac{\,dy}{\,dx}+\frac{2}{x}y=\left(\frac{1}{x}+\  frac{2}{x^2}\right)e^x

    Now apply the technique of the integrating factor. Can you continue from here?

    (b) Find the solution of

    \frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}

    given that at x=0,y=2 and \frac{dy}{dx}=4
    The characteristic equation is r^2-4r+5=0 which means that r=2\pm i (Verify)

    Thus, the homogeneous solution is y_c=e^{2x}\left[c_1\sin\left(x\right)+c_2\cos\left(x\right)\right]

    Now apply the method of undetermined coefficients to determine the particular solution. Once you find the general solution (homogeneous + particular), apply the initial conditions. Can you take it from here?

    3 (a) Find the following two integrals:
    \int\frac{x}{1-x^2}dx and \int\frac{x}{1-x^2}dx
    They're the same thing....

    Apply the substitution z=1-x^2...

    Find y in terms of x, givene that
    \frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}

    and that y =1 when x = 0.
    You should do this with no problem once you have figured out to integrate \int\frac{x}{1-x^2}\,dx. Again, you're using the technique of the integrating factor to solve this.

    (b) Solve the differential equation
    \frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0

    given that y=1 and \frac{dy}{dx}=3 when x=0.
    The corresponding characteristic equation is r^2-2r+2=0 which yields r=1\pm i (Verify)

    The general solution would be y=e^x\left[c_1\sin\left(x\right)+c_2\cos\left(x\right)\right]

    Can you take it from here and apply the initial conditions?

    Does this make sense?
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  4. #4
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    Quote Originally Posted by Lonehwolf View Post
    1 Obtain the General Solution of the differential equations:
    (i) x\frac{dy}{dx} + y = xe^{3x}

    (ii) \frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}

    2 (a) Find the general solution of the differential equation
    x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x

    (b) Find the solution of

    \frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}

    given that at x=0,y=2 and \frac{dy}{dx}=4

    3 (a) Find the following two integrals:
    \int\frac{x}{1-x^2}dx and \int\frac{x}{1-x^2}dx

    Find y in terms of x, givene that
    \frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}

    and that y =1 when x = 0.

    (b) Solve the differential equation
    \frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0

    given that y=1 and \frac{dy}{dx}=3 when x=0.
    2. a) This is a first order linear ODE.

    So we use the Integrating Factor method.

    Divide everything through by x first, to get

    \frac{dy}{dx} + \frac{2}{x}y = \left(\frac{1}{x} + \frac{2}{x^2}\right)e^x.

    Now we find the Integrating Factor and multiply both sides of the equation by it.

    It's e^{\int{\frac{2}{x}\,dx}}=e^{2\ln{x}}=e^{\ln{x^2}}  =x^2.

    So the DE becomes

    x^2\frac{dy}{dx} + 2xy = (x + 2)e^x.

    The left hand side is the product rule expansion of \frac{d}{dx}(x^2y)

    So \frac{d}{dx}(x^2y) = (x + 2)e^x

    x^2y = \int{(x + 2)e^x\,dx}.

    Once again, the RHS can be evaluated using Integration by Parts.
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  5. #5
    Junior Member Lonehwolf's Avatar
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    Thanks for the replies. Regarding the - and + it was - if I remember correctly, and those weren't supposed to be the same thing >.< There was some minor difference.

    Alas, I woke up late and couldn't get these done so... i'm as good as roasted, but that's completely my fault for not switching ISPs and for being a blockhead >.>

    Thanks for all the inputs, and Pivot thanks for explaining Integrating Factor, I was lost with those for the time being. I may get yelled at but at least I understood 'em xD
    Last edited by Lonehwolf; December 23rd 2008 at 10:38 PM.
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