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Thread: linear differential equations

  1. #1
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    linear differential equations

    show that the differential eqn x^2 * dy/dx -y = 2x^3*e^(-1/x)
    is linear, and solve it using the integrating factor method.

    I have no idea how to even get started on this problem, if someone could show me in a few steps it would really help me out. I have alot more of these to do.
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  2. #2
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    Quote Originally Posted by samdmansam View Post
    show that the differential eqn x^2 * dy/dx -y = 2x^3*e^(-1/x)
    is linear, and solve it using the integrating factor method.

    I have no idea how to even get started on this problem, if someone could show me in a few steps it would really help me out. I have alot more of these to do.
    $\displaystyle y' - \frac{1}{x^2}y = 2xe^{-1/x}$, let $\displaystyle p(x) = e^{\int -\frac{1}{x^2} dx}$.

    The solution is given by $\displaystyle y = \frac{1}{p(x)} \left\{ \int 2xp(x)e^{-1/x} dx + C \right\}$.
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  3. #3
    Super Member PaulRS's Avatar
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    Ok, suppose you have a differential equation of the form: $\displaystyle f'(x)+a(x)\cdot{f(x)}=b(x)$

    Let$\displaystyle f(x)\cdot{e^{\int_c^xa(t)dt}}=q(x)$ (1) be an auxiliary function then: $\displaystyle f'(x)\cdot{e^{\int_c^xa(t)dt}}+a(x)\cdot{f(x)}\cdo t{e^{\int_c^xa(t)dt}}=q'(x)$

    But note that: $\displaystyle \left(f'(x)+a(x)\cdot{f(x)}\right)\cdot{e^{\int_c^ xa(t)dt}}=b(x)\cdot{e^{\int_c^xa(t)dt}}=q'(x)$

    Thus, by the FTC: $\displaystyle \int_c^xb(z)\cdot{e^{\int_c^za(t)dt}}dz=q(x)-q(c)=q(x)-f(c)$ check in (1) that $\displaystyle q(c)=f(c)$

    Thus we have by (1) that: $\displaystyle e^{-\int_c^xa(t)dt}\cdot{\int_c^xb(z)\cdot{e^{\int_c^z a(t)dt}}dz}+f(c)\cdot{e^{-\int_c^xa(t)dt}}=f(x)$

    This is the general solution to the first order equation

    So suppose you are given $\displaystyle f(0)$ for example, then you should let $\displaystyle c=0$ above
    Last edited by PaulRS; Jun 4th 2008 at 05:03 AM.
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