Solving Math Problems (Part 17)

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

Section 1.4, Problem 14

Solve the following initial value problem:

xyy' = 2y² + 4 x², y(2) = 4

Dividng by x² we get:

(y/x) y' = 2(y/x)² + 4

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

u(u + u'x) = 2u² + 4

u² + uu'x = 2u² + 4

uu' / (u² + 4) = 1 / x

Integrating both sides we get:

We now make another change of variable to z such that z = u² + 4 and dz = 2u du.

½ln |z| = ln |x| + C

Changing back to u from z gives

½ln |u² + 4| = ln |x| + C

Taking e to the power of both sides gives:

Re-arranging to isolate u gives:

And finally changing back to y from u we have:

Now using the initial condition, we find that C = 2 and thus the specific function that satisfies the differential equation with this initial condition is:

Section 1.4, Problem 15

Find the function that satisfies:

Rearranging we get:

yy' - xy' = y + x

Dividing by x gives:

(y/x) y' - y' = (y/x) + 1

Now changing variable with u = (y/x) and y' = u + u'x

u (u + u'x) - (u + u'x) = u + 1

u² + uu'x - u - u'x = u + 1

uu'x - u'x = 1 + 2u - u²

u'x(u - 1) = 1 + 2u - u²

We make another change of variable such that z = 1 + 2u - u² and dz = (2 - 2u) du and we get:

Integrating both sides gives:



½ln|z| = ln|x| + C

z = Cx²

Changing back to u from z:

1 + 2u - u² = Cx²

Changing back to y from u:

x² + 2xy - y² = C

Then using the initial condition, we find that C = -4