{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

55_pdfsam_math 54 differential equation solutions odd

# 55_pdfsam_math 54 differential equation solutions odd - = 1...

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

Exercises 2.3 Table 2–D : Euler’s method approximations for the solution of y + y 1 + sin 2 x = x , y (0) = 2, at x = 1 with h = 0 . 1. k x k y k k x k y k k x k y k 0 0.0 2.0000 4 0.4 1.3584 8 0.8 1.0304 1 0.1 1.8000 5 0.5 1.2526 9 0.9 0.9836 2 0.2 1.6291 6 0.6 1.1637 10 1.0 0.9486 3 0.3 1.4830 7 0.7 1.0900 Table 2–E : Euler’s method approximations for the solution of y + y 1 + sin 2 x = x , y (0) = 2, at x = 1 with h = 0 . 05. n x n y n n x n y n n x n y n 0 0.00 2.0000 7 0.35 1.4368 14 0.70 1.1144 1 0.05 1.9000 8 0.40 1.3784 15 0.75 1.0831 2 0.10 1.8074 9 0.45 1.3244 16 0.80 1.0551 3 0.15 1.7216 10 0.50 1.2747 17 0.85 1.0301 4 0.20 1.6420 11 0.55 1.2290 18 0.90 1.0082 5 0.25 1.5683 12 0.60 1.1872 19 0.95 0.9892 6 0.30 1.5000 13 0.65 1.1490 20 1.00 0.9729 29. In the presented form, the equation
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 1 e 4 y + 2 x is, clearly, not linear. But, if we switch the roles of variables and consider y as the independent variable and x as the dependent variable (using the connection between derivatives of inverse functions, that is, the formula y ( x ) = 1 /x ( y )), then the equation transforms to dx dy = e 4 y + 2 x ⇒ dx dy − 2 x = e 4 y . This is a linear equation with P ( y ) = − 2. Thus the integrating factor is µ ( y ) = exp ±Z ( − 2) dy ² = e − 2 y 51...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online