{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ordinary Diff Eq Exam Review Solutions 20

# Ordinary Diff Eq Exam Review Solutions 20 - 22 1 Solutions...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 22 1 Solutions 16. The given equation is in standard form, , 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 . Thus and taking antiderivatives of both sides gives where is a constant. Now multiply by to get for the general solution. 17. The given equation is in standard form, , tive is , and the integrating factor is by the integrating factor to get of which is a perfect derivative . Thus antiderivatives of both sides gives where is a constant. Now multiply by , an antideriva. Now multiply , the left hand side and taking to get for the general solution. 18. The given diﬀerential equation is in standard form, , 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 . Thus . Now assume . Then taking antiderivatives of both sides gives where is a constant. Now multiply by to get for the general solution. If antiderivatives leads to and hence the general solution in this case. then taking is 19. In standard form we get . In this case , 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 . Thus and taking antiderivatives of both sides gives where is a constant. Now multiply by and we get for the general solution. 20. Divide by to put the equation in the standard form In this case grating factor is get , an antiderivative is , and the inte. Now multiply by the integrating factor to ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online