anonymous
  • anonymous
\[3y'+2y=x^3sin(\pi*x)\]
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Give a min plzz, i Think we need to use the integrating factor here
anonymous
  • anonymous
We can write the equation as - \[y ^{'} + (2/3)y = (1/3)x ^{2}\sin \pi x\] Hence the integrating factor becomes \[\exp[\int\limits_{}^{}(2/3)dx] = e ^{(2/3)x}\] Multiplying both sides by this factor of the equation we get - \[e ^{(2/3)x}y {^'} + (2/3)ye ^{(2/3)x} = (1/3)e ^{(2/3)x}x ^{2}\sin \pi x\] Now the left hand side is the derivative of
anonymous
  • anonymous
Continuing from earlier post - Left hand side is \[d/dx(ye ^{(2/3)x})\] Hence the solution becomes - \[ye ^{(2/3)x} = (1/3)\int\limits_{}^{}e ^{(2/3)x} x ^{2} \sin \pi x dx\] Right hand side we need to use Integration by parts, using x^2 as first function, sin pi x as the seond function and the exponential as the third function

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anonymous
  • anonymous
Can you do the integration by parts?
anonymous
  • anonymous
When I use the method of udetermined coefficients I get; \[\huge y_g = c_1e^{-\frac{2}{3}x}+\left(\frac{1}{2}x^3-\frac{9}{4}x^2+\frac{27}{4}x-\frac{81}{8} \right) \]\[\huge*\left(\frac{2}{9\pi^2+4}\text{sin}(\pi*x)-\frac{3\pi}{9\pi^2+4}\text{cos}(\pi*x) \right)\]

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