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Ordinary Diff Eq Exam Review Solutions 30

# Ordinary Diff Eq Exam Review Solutions 30 - 32 1 Solutions...

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Unformatted text preview: 32 1 Solutions and solve for . The initial condition implies to get and so . The solution is thus . 10. Note that is a solution. Divide by to get . Let . Then and substituting gives . In the standard form for linear equations this becomes . Multiplying by the integrating factor gives so that . Hence . Now go back to the original function by solving for . Thus . The general solution is and . 11. Note that is a solution. First divide both sides by . Let . Then , so . Substituting , which in standard form is gives to get . An integrating factor is so that . Integrating both sides gives , where the integral of the right hand side is done by the substitution . Solving for gives . Since we ﬁnd 12. Note that . is a solution. First divide both sides by . Let . Then stituting gives , so to get . Sub- , which in standard form is . An integrating factor is Thus Integrating both sides gives , where the integral of the right hand side is computed using integration by parts with and . Solving for gives . Since we ﬁnd . 13. Note that is a solution. Divide by . Let gives . Then and to get and substituting . In standard form we get . Multiplying by the integrating factor ...
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