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  1. MHB Apprentice

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    #1
    solve the following differential equation with the suggested change of variables.


  2. Indicium Physicus
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    #2
    What progress have you made so far on these DE's? Do you know how to start solving them?

  3. MHB Apprentice

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    #3 Thread Author
    actually i don't know how to solve them.Can anyone help please.

  4. Indicium Physicus
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    #4
    Sure! So on the first one, the suggested substitution is $x=e^t$. The reason this works is because the original DE is a Cauchy-Euler equation, where you're multiplying each derivative by successively lower powers of $x$. So, if we do this substitution, we need all the derivatives in place (note here that $y'=dy/dx$ and $\dot{y}=dy/dt$):
    \begin{align*}
    x&=e^t \\
    \frac{dx}{dt}&=e^t \\
    \frac{dt}{dx}&=e^{-t} \\
    y'&=\frac{dy}{dx}=\frac{dy}{dt} \, \frac{dt}{dx} = \dot{y} \, e^{-t} \\
    y''&=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} \, \frac{dy}{dx} = \left[\frac{d}{dt} \, \frac{dy}{dx}\right] \frac{dt}{dx}
    =\frac{d}{dt}\left[ \dot{y} \, e^{-t} \right] e^{-t}
    =\left[ \ddot{y} e^{-t} - \dot{y} e^{-t} \right] e^{-t} = e^{-2t} (\ddot{y}-\dot{y}).
    \end{align*}
    Now, substitute all this stuff into the original DE. What do you get?

  5. Pessimist Singularitarian
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    #5
    For the second problem, we are given:

    $ \displaystyle \left(1+x^2\right)^2\frac{d^2y}{dx^2}+2x\left(1+x^2\right)\frac{dy}{dx}+my=0$

    We are told to use the substitutions:

    $ \displaystyle x=\tan(t)\implies \frac{dx}{dt}=\sec^2(t)$

    $ \displaystyle y(x)=z(t)$

    In the second substitution, if we implicitly differentiate w.r.t $t$, we obtain:

    $ \displaystyle \frac{dy}{dx}\cdot\frac{dx}{dt}=\frac{dz}{dt}$

    Using the implication from the first substitution, there results:

    $ \displaystyle \frac{dy}{dx}=\cos^2(t)\frac{dz}{dt}$

    Now, if we differentiate again w.r.t $t$, we have:

    $ \displaystyle \frac{d^2y}{dx^2}\cdot\frac{dx}{dt}=\cos^2(t)\frac{d^2z}{dt^2}-2\cos(t)\sin(t)\frac{dz}{dt}$

    Again, using the implication from the first substitution, there results:

    $ \displaystyle \frac{d^2y}{dx^2}=\cos^4(t)\frac{d^2z}{dt^2}-2\cos^3(t)\sin(t)\frac{dz}{dt}$

    Now, let's plug all of the substitutions and their implications into the given ODE:

    $ \displaystyle \left(1+\tan^2(t)\right)^2\left(\cos^4(t)\frac{d^2z}{dt^2}-2\cos^3(t)\sin(t)\frac{dz}{dt}\right)+2\tan(t)\left(1+\tan^2(t)\right)\left(\cos^2(t)\frac{dz}{dt}\right)+mz=0$

    Next, let's use the Pythagorean identity $1+\tan^2(\theta)=\sec^2(\theta)$:

    $ \displaystyle \sec^4(t)\left(\cos^4(t)\frac{d^2z}{dt^2}-2\cos^3(t)\sin(t)\frac{dz}{dt}\right)+2\tan(t)\sec^2(t)\left(\cos^2(t)\frac{dz}{dt}\right)+mz=0$

    Distribute:

    $ \displaystyle \frac{d^2z}{dt^2}-2\tan(t)\frac{dz}{dt}+2\tan(t)\frac{dz}{dt}+mz=0$

    Combine like terms:

    $ \displaystyle \frac{d^2z}{dt^2}+mz=0$

    Can you proceed?

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    #6
    Just to follow up with the second problem, givent that $m$ is a positive real number, we see the roots of the characteristic, or auxiliary equation are:

    $ \displaystyle r=\pm\sqrt{m}i$

    And so, by the theory of linear homogeneous equations, we know the general solution will be the two-parameter family:

    $ \displaystyle z(t)=c_1\cos(\sqrt{m}t)+c_2\sin(\sqrt{m}t)$

    Back-substituting for $z$ and $t$, we obtain:

    $ \displaystyle y(x)=c_1\cos\left(\sqrt{m}\arctan(x)\right)+c_2\sin\left(\sqrt{m}\arctan(x)\right)$

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