Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A mathematics club studies the family of curves given by the equation for . The equation is not exact as it stands, but one factor that depends on alone makes it exact. A member finds the solution curves in the form and picks the curve through . Find the integrating factor at , the constant to four decimal places, and the positive value of on the curve at , to four decimal places.
is exact when , and then with and . If depends on only, . If depends on only, . For homogeneous , use .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. , , and the curve through .
Plan. Find the integrating factor, confirm the solution, then use the point and the value .
Step 1. The clue. and , so . The factor is . At it equals .
Step 2. The new equation. has . Integrating gives , and .
Step 3. The constant. At : .
Step 4. At . . This gives , so and .
Answer: the factor is at , and .
Watch out. The factor is a function of , so it is and not the number or .
Log in to join the discussion.
Loading…