Solving Math Problems (Part 7)

More problems from Advanced Engineering Mathematics, 6th Edition by Kreyszig.

Section 1.2, Problem 21

Solve the initial value problem

y' = x / y given y(2) = 0

Separating the variables gives:

y y' = x

Integrating both sides gives:

∫ y dy = ∫ dx

½ y² = x + C

Using the initial condition, we find that C = -2 so that the solution is:

Section 1.2, Problem 22

Solve the initial value problem

y y' = sin² x given y(0) = ⎷3

Since the variables are already separated, we shall integrate both sides

∫ y dy = ∫ sin² x dx

½ y² = ½ x - ¼ sin 2x + C

Using the initial condition we find that 2C = 3 so that the final solution is:

Section 1.2, Problem 23

Solve the initial value problem

dr sin 𝛉 = 2r cos 𝛉 d𝛉 given r(𝛑/4) = -2

Separating variables gives:

Integrating both sides give:

½ ln | r | = ln | sin 𝛉 | + C

Taking e to the power of both sides, we get:

r = C sin² 𝛉

Substituting the given initial condition and solving for C we find that C = -4  thus the solution is:

r = -4 sin² 𝛉