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Thread: 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) $\displaystyle x\frac{dy}{dx} + y = xe^{3x}$

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

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

    (b) Find the solution of

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

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

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

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

    and that y =1 when x = 0.

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

    given that y=1 and $\displaystyle \frac{dy}{dx}=3$ 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) $\displaystyle x\frac{dy}{dx} + y = xe^{3x}$

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

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

    (b) Find the solution of

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

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


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

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

    and that y =1 when x = 0.

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

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

    $\displaystyle \frac{d}{dx}(xy)$

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

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

    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) $\displaystyle x\frac{dy}{dx} + y = xe^{3x}$
    Divide through by the coefficient of $\displaystyle \frac{\,dy}{\,dx}$ to get $\displaystyle \frac{\,dy}{\,dx}+\frac{1}{x}y=e^{3x}$

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

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

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

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

    (b) Find the solution of

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

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

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

    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:
    $\displaystyle \int\frac{x}{1-x^2}dx$ and $\displaystyle \int\frac{x}{1-x^2}dx$
    They're the same thing....

    Apply the substitution $\displaystyle z=1-x^2$...

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

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

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

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

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

    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) $\displaystyle x\frac{dy}{dx} + y = xe^{3x}$

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

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

    (b) Find the solution of

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

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

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

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

    and that y =1 when x = 0.

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

    given that y=1 and $\displaystyle \frac{dy}{dx}=3$ 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

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

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

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

    So the DE becomes

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

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

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

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

    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; Dec 23rd 2008 at 10:38 PM.
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