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Monday, October 14, 2019

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 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