# Q1ts - the standard procedure in standard form this is dy...

This preview shows pages 1–2. Sign up to view the full content.

Quiz I 1. Find the solution to 2 y dy dx - x = xy 2 satisfying y (0) = - 1. Solution. This equation can be rewritten as dy dx = x ( y 2 + 1 2 y ) Note that this has no equilibrium solutions, and it is separable. So, following the standard procedure Z 2 y 1 + y 2 dy = Z xdx, and ln(1 + y 2 ) = x 2 / 2 + C. If y (0) = - 1, then C = ln2. So y 2 + 1 = e x 2 / 2+ln 2 = 2 e x 2 / 2 and y = - p 2 e x 2 / 2 - 1 . Note that you need the minus sign to make y (0) = - 1.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 2. Find the general solution to (1 + x ) dy dx + y = cos x . Solution . You may have noticed that this equation has already been multiplied by an integrating factor, and been able to skip some steps below. If not, following
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the standard procedure, in standard form this is dy dx + 1 1 + x y = cos x 1 + x . So a ( x ) = 1 / (1 + x ), A ( x ) = R a ( x ) dx = ln | 1 + x | and the integrating factor is e A = | 1+ x | = ± (1+ x ). Either one will work. Using +(1+ x ), you get back to the original equation! So you have (1 + x ) dy dx + y = cos x, d dx (1 + x ) y = cos x, (1 + x ) y = sin x + C. So the general solution is y ( x ) = sin x 1 + x + C 1 + x ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

Q1ts - the standard procedure in standard form this is dy...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online