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    #1
    $\tiny{206.q3.2}\\$
    $\textsf{3. use the method of integrating factor}\\$
    $\textsf{to find the general solution to the first order linear differential equation}\\$
    \begin{align}
    \displaystyle
    \frac{dy}{dx}+5y=10x
    \end{align}
    $\textit{clueless !!!}$
    Last edited by karush; December 10th, 2016 at 20:59.

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    #2
    Given a first order linear ODE of the form:

    $ \displaystyle \d{y}{x}+f(x)y=g(x)$

    We can use an integrating factor $\mu(x)$ to make the LHS of the ODE into the derivative of a product, using the special properties of the exponential function with regard to differentiation. Consider what happens if we multiply though by:

    $ \displaystyle \mu(x)=\exp\left(\int f(x)\,dx\right)$

    We get:

    $ \displaystyle \exp\left(\int f(x)\,dx\right)\d{y}{x}+\exp\left(\int f(x)\,dx\right)f(x)y=g(x)\exp\left(\int f(x)\,dx\right)$

    Now, let's use:

    $ \displaystyle F(x)=\int f(x)\,dx\implies F'(x)=f(x)$

    And we now have:

    $ \displaystyle \exp\left(F(x)\right)\d{y}{x}+\exp\left(F(x)\right)F'(x)y=g(x)\exp\left(F(x)\right)$

    Now, if we observe that, via the product rule, we have:

    $ \displaystyle \frac{d}{dx}\left(\exp(F(x))y\right)=\exp\left(F(x)\right)\d{y}{x}+\exp\left(F(x)\right)F'(x)y$

    Then, we may now write our ODE as:

    $ \displaystyle \frac{d}{dx}\left(\exp(F(x))y\right)=g(x)\exp\left(F(x)\right)$

    Now, we may integrate both sides w.r.t $x$.

    So, in the given ODE:

    $ \displaystyle \d{y}{x}+5y=10x$

    We identify: $ \displaystyle f(x)=5$

    And so we compute the integrating factor as:

    $ \displaystyle \mu(x)=\exp\left(5\int\,dx\right)=$?

  3. MHB Master
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    #3 Thread Author
    wow that was a great help. but I'll have go thru this tonmorro ...

    ok, got it, but have never come up with that😎
    Last edited by karush; December 10th, 2016 at 21:15.

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