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Math Help - Solution of 4t dy/dt + y = t^3

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    s3a
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    Solution of 4t dy/dt + y = t^3

    The question is attached (along with the answer). It's asking me to guess the solution of this differential equation. I think I would have to use the reduction of order method if I were given a particular solution but I am not given one. I'm assuming there is a logical way to guess this rather than just randomly guessing in a brute force manner. Could someone explain to me what the thought process is for this?

    Thanks in advance!
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    Re: Solution of 4t dy/dt + y = t^3

    Quote Originally Posted by s3a View Post
    The question is attached (along with the answer). It's asking me to guess the solution of this differential equation. I think I would have to use the reduction of order method if I were given a particular solution but I am not given one. I'm assuming there is a logical way to guess this rather than just randomly guessing in a brute force manner. Could someone explain to me what the thought process is for this?

    Thanks in advance!
    Am not sure how helpful this will be but the problem is of the type Cauchy-Euler

    Cauchy

    This will at leat give you something to look up.

    I will also give two ideas:

    Idea 1: Solve the homogenous problem first.

    4ty'+y=0 \iff \frac{dy}{y}=-\frac{1}{4}\frac{dt}{t} \implies \ln(y)=-\frac{1}{4}\ln(t)+C \implies y=\frac{c}{t^{1/4}}

    Now find the particular solution by guessing it is a Monomial of degree 3. (do you see what it cannot have any lower degree terms.)

    Idea 2
    Let y=f(u), u=\ln(t)

    Then \frac{dy}{dt}=\frac{df}{du}\frac{du}{dt}=\frac{1}{  t}\frac{df}{du}

    This will transform the equation into

    4t\left( \frac{1}{t}\frac{df}{du}\right)+f(u)=e^{3u} \iff 4f'+f=e^{3u}

    This is a first order linear differential equation with contant coeffiencts.

    Just FYI the usual "guess" is y=At^n
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