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anonymous

  • 4 years ago

Solve the differential equation \[ y'+3y-2=4cos(\pi*x)+x^2-4e^{2x}\] using the method of undetermined coefficients

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  1. TuringTest
    • 4 years ago
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    \[y'+3y=4\cos(\pi x)+x^2-4e^{2x}+2\]now it's linear and should be doable with an integrating factor

  2. TuringTest
    • 4 years ago
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    \[\mu(x)=e^{\int 3dx}=e^{3x}\]multiply through...

  3. TuringTest
    • 4 years ago
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    \[(ye^{3x})'=4e^{3x}\cos(\pi x)+x^2e^{3x}+4e^{5x}+2e^{3x}\]\[\int (ye^{3x})'=ye^{3x}+C=\int4e^{3x}\cos(\pi x)+x^2e^{3x}+4e^{5x}+2e^{3x}dx\]some parts of this integral may get a little ugly, but nothing integration by parts can't handle

  4. TuringTest
    • 4 years ago
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    if you're not familiar with what I did on the left side you should acquaint yourself with solving linear equations with integrating factors

  5. anonymous
    • 4 years ago
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    USING THE METHOD OF UNDETERMINED COEFFICIENTS. I know how to use the integrating factor method and I hate integration by parts.

  6. anonymous
    • 4 years ago
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    Hatred is foolish. I distaste integrating by parts so please use the method of undetermined coefficients.

  7. TuringTest
    • 4 years ago
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    Undetermined coefficients for a first-order DE? never heard of such a thing

  8. TuringTest
    • 4 years ago
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    well this was interesting, I didn't know you could use undetermined coefficients here, but here it is:

  9. TuringTest
    • 4 years ago
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    \[y'+3y=4\cos(\pi x)+x^2-4e^{2x}+2\]\[y_c=c_1e^{-3}\]\[Y_p=A\cos(\pi x)+B\sin(\pi x)+Ce^{2x}+Dx^2+Ex+F\]\[Y'_p=-A\pi\sin(\pi x)+B\pi\cos(\pi x)+2Ce^{2x}+2Dx+E\]plugging this into the equation we find that matching the coefficients on the left and right for\[Y_p'+3Y_p\]leads to the system\[-\pi A+3B=0\]\[3A+B\pi=4\]\[5C=-4\]\[3D=1\]\[3E+2D=0\]\[E+3F=2\]C, D, E, and F are found easily, and A and B can be found quickly using Cramer's rule. We find that\[A=\frac{12}{\pi^2+9}\]\[B=\frac{4\pi}{\pi^2+9}\]\[C=-\frac54\]\[D=\frac13\]\[E=-\frac23\]\[F=\frac{20}{27}\]so our solution is\[y(x)=c_1y_c+Y_p=\dots\]\[c_1e^{-3x}+\frac{12}{\pi^2+9}\cos(\pi x)+\frac{4\pi}{\pi^2+9}\sin(\pi x)-\frac45e^{2x}+\frac13x^2-\frac29x+\frac{20}{27}\]actually that wasn't so bad. maybe even easier than integration by parts. Interesting!

  10. TuringTest
    • 4 years ago
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    *the last number is\[+\frac{20}{27}\]in case you can't see it

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