Solving Math Problems (part 16)

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

Section 1.4, Problem 11

Solve the initial value problem:

xy' = x + y, y(1) = -7.4

Since we solved the general solution in Problem 1 and found it to be y = x( ln |x| + C), we will now simply apply the initial condition.

 -7.4 = 1 ( ln | 1 | + C )

∴ C = -7.4

The specific curve is described by:

y = x ( ln |x| -7.4 )

Section 1.5, Problem 12

Solve the initial value problem:

xy' = 2x +2y, y(0.5) = 0

From here, we find the general solution is:

y = C x² - 2x

Substituting the given initial value, we get:

0 = C (0.25) - 2 (0.5)

∴ C = 4

The specific curve is described by:

y = 4x² - 2x

Section 1.5, Problem 13

Solve the initial value problem:

yy' = x³ + y²/x, y(2) = 6

Dividing by x gives:

(y/x) y' = x² + y²/x²

Changing the variable by u = y/x and y' = u + u'x, we get:

u ( u + u'x ) = x² + u²

Separating the variables and integrating gives:

∫u du = ∫x dx

½u² = ½x² + C

Converting to y gives:

Using the given initial value, we have:

∴ C = 5

The specific solution is