428-MTH321:-MTH321:

︠853a39b3-76c2-48f1-aed6-e795b28927bai︠
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<h1>MTH321: &nbsp;Homework Ten</h1>

<h2>1. &nbsp;Section 3.1.1, #5</h2>
<p>Find a particular solution to the nonhomogeneous equation</p>
<p>$$u'' + u' + 2u = \sin^2(t)$$</p>

<p>Solution: &nbsp;rewrite the right hand side using the double angle formula</p>
<p>$$\sin^2(t) &nbsp; = \frac{1}{2} - \frac{\cos(2t)}{2}$$</p>
<p>Then we make a guess for the particular solution $u_p = A + B\sin(2t) + C\cos(2t)$ and plug in to solve for the coefficients. &nbsp;The result is</p>
<p>$$ u_p = \frac{1}{4} - \frac{1}{8}\cos(2t) + \frac{1}{8}\sin(2t)$$</p>

︡4071e1b1-3e69-42ec-8544-017540e76b5a︡{"html": "<h1>MTH321: &nbsp;Homework Ten</h1>\n\n<h2>1. &nbsp;Section 3.1.1, #5</h2>\n<p>Find a particular solution to the nonhomogeneous equation</p>\n<p>$$u'' + u' + 2u = \\sin^2(t)$$</p>\n\n<p>Solution: &nbsp;rewrite the right hand side using the double angle formula</p>\n<p>$$\\sin^2(t) &nbsp; = \\frac{1}{2} - \\frac{\\cos(2t)}{2}$$</p>\n<p>Then we make a guess for the particular solution $u_p = A + B\\sin(2t) + C\\cos(2t)$ and plug in to solve for the coefficients. &nbsp;The result is</p>\n<p>$$ u_p = \\frac{1}{4} - \\frac{1}{8}\\cos(2t) + \\frac{1}{8}\\sin(2t)$$</p>"}︡