Solving Math Problems (Part 8)

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

Section 1.2, Problem 24

Solve the initial value problem:

dr/dt = -2tr with r(0) = 4.6

Separating variables gives us:

Integrating both sides gives:

ln | r | = - t² + C

Putting e to the power of both sides gives

Using the initial condition, we find that C = 4.6 so the solution is:

Section 1.2, Problem 25

Solve the initial value problem:

y' = y tanh x with y(0) = 1.4

Separating variables gives:

Integrating both sides gives:

Looking up the right hand side in Jan J. Tuma's Engineering Mathematics Handbook, 3rd Edition, Section 19, table 81, we have

ln | y | = ln | cosh x | + C

Raising e to the power of both sides gives:

y = C cosh x

Using the initial conditions and solving for C, we find C = 1.4 so the solution is:

y = 1.4 cosh x

Section 1.2, Problem 26

Solve the initial value problem:

y³ y' + x³ = 0 with y(2) = 0

Separating variables we have:

y³ y' = - x³

Integrating both sides gives:

∫ y³ dy = ∫ - x³ dx

¼y⁴= - ¼x⁴+ C

Using the initial condition, we can solve for C and find C = 4.  Some re-arrangement gives us the solution of:

y⁴+ x⁴= 16