Ordinary Diff Eq Exam Review Solutions 18

Ordinary Diff Eq Exam Review Solutions 18 - 20 1 Solutions...

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Unformatted text preview: 20 1 Solutions , . Now integrate both sides to get where we computed by parts two times. Dividing by gives . 9. In this case Now multiply to get to twice. Dividing by and the integrating factor is . , which simplifies . Now integrate both sides to get , where we computed by parts gives . 10. In standard form this equation becomes Using partial fractions we get , an antiderivative is , and the integrating factor is . Now multiply by the integrating factor to get the left hand side of which is a perfect derivative and taking antiderivatives of both sides gives is a constant. Now multiply by to get general solution. . Thus where for the 11. In standard form we get . An integrating factor is Thus Integrating both sides gives , where the integral of the right hand side is done by parts. Now divide by the integrating factor to get . 12. The given differential equation is in standard form, tiderivative is , and the integrating factor is multiply by the integrating factor to get the left hand side of which is a perfect derivative If then taking antiderivatives of both sides gives where is a constant. Now multiply by to get the general solution. In the case then and , an an. Now . Thus for . ...
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