## anonymous 4 years ago $3y'+2y=x^3sin(\pi*x)$

1. anonymous

Give a min plzz, i Think we need to use the integrating factor here

2. 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

3. 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

4. anonymous

Can you do the integration by parts?

5. 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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