Suppose a friend tries to solve the differential equation dy/dt = yt by integrat

Question

Suppose a friend tries to solve the differential equation dy/dt = yt by integrating to get y=1/2yt2+ C. a) Solve y=1/2yt2+ C to get a formula for "y" in terms of "t", and show that this is NOT a solution of the differential equation. b) What is the friend's mistake? c) What is the correct general solution?
Suppose a friend tries to solve the differential equation dy/dt = yt by integrating to get y=1/2yt2+ C. a) Solve y=1/2yt2+ C to get a formula for "y" in terms of "t", and show that this is NOT a solution of the differential equation. b) What is the friend's mistake? c) What is the correct general solution?

Explanation / Answer

a) Solve y=1/2yt2+ C to get a formula for "y" in terms of "t So, y-1/2*y*t^2 = C or y(1-1/2*t^2)= C or y = C/(1-1/2t^2) Does dy/dt = yt ? Check dy/dt = -C(1-1/2t^2)^-2*(-2t) = 2Ct/(1-1/2t^2)^2 yt = Ct/(1-1/2t^2) So, dy/dt not equal to yt b) Friend mistake is that he forgot y is a function of t c) dy/dt = yt or dy/y = t dt Integrate to get ln(y) = 1/2 * t^2 + C

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.