Solving Math Problems (Part 13)

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

Section 1.4, Problem 7

Find the general solution of

xy' = y + x² sec (y/x)

Dividing by x gives:

y' = (y/x) + x sec (y/x)

Substituting u = y/x and y' = u + u'x we get:

u + u'x = u + x sec u

Re-arranging we get:

∫cos u du = ∫ dx

sin u = x + C

u = arc sin (x + C)

y = x arc sin (x + C)

Section 1.4, Problem 8

Find the general solution of

Substituting u = y/x and y' = u + u'x we get:

u(u + u'x) = u² + x/(x²+1)

Simplifying we get:

uu' = 1/(x² + 1)

Integrating both sides, we have:

u²/2 = arc tan x + C

Substituting y = ux, we have:

Solving Math Problems (Part 12)

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

Section 1.4, Problem 5

Find the general solution of:

x²y' = x² - xy + y²

Dividing by x² we get

Changing the variable using u = y/x and y' = u + u'x we get

u + u'x = 1 - u + u²

Rearranging we get:

Integrating both sides:



Substituting back y instead of u, we have:

Section 1.4, Problem 6

Find the general solution of:

xy' - y = 3x⁴cos²( y / x )

Rearranging we get:

y' - (y/x) = 3x³ cos²( y / x )

Changing the variable using u = y/x and y' = u + u'x we get

u + u'x - u = 3x³ cos² u

Separating the variables we have:

Integrating both sides we have:

tan u = x³ + C

u = arc tan (x³ + C)

Changing back to y gives:

y = x arc tan (x³ + C)

Solving Math Problems (Part 11)

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

Section 1.4, problem 3

Find the general solution to:

xy' = (y - x)³ + y

Expanding the bracket and dividing everything by x³ we get:

Using a change of variable where u = y/x and y' = u + u'x we get:

Simplifying we get:

Separating the variables by cross multiplication and then integrating both sides we have:

Some re-arrangement is necessary before we can convert back to y from u.

Converting back to y from u we have:

Section 1.4, Problem 4

Find the general solution of:

xy' = y² + y

Dividing by x² we get:

Change of variable using u = y/x and y' = u + u'x gives:

This simplifies to:

u' = u²

Integrating, we get:

 Re-arranging:

 Changing back to y from u gives



Solving Math Problems (part 10)

More problems from Advanced Engineering Mathematics, 6th Edition by Kreyszig.  Section 1.2 has 2 more problems; #29 is asking for a proof and #30 asks to use this proof to re-work an example given in the book.  Following that, section 1.3 is about modeling and the exercises at the end are essentially physics word problems wherein the answer involves creating a differential equation and then solving an initial value problem.  I am skipping all that to go to section 1.4 which involves a change of variable to convert a given differential equation into a separable equation.

Section 1.4, problem 1

Find the general solution of:

xy' = x + y

We can re-arrange this to be:

Using a change of variable such that u = y/x and y' = u + u'x we get:

u + u'x = 1 + u

With some re-arrangement we get:

Integrating:

gives:

u = ln |x| + C

Changing back to y using y = ux, we get:

y = x (ln |x| +C)

Section 1.4, problem 2

Find the general solution of:

xy' - 2y = 3x

Re-arrangement gives:

Using a change of variable such that u = y/x and y' = u + u'x we get:

u + u'x - 2u = 3

Re-arranging to separate the variables we have:

Integrating both sides we get:

Solving:

ln |u+3| = ln |x| + C

Raising e to the power of both sides gives:

u + 3 = Cx

u = Cx - 3

Changing back to y using y = ux, we get:

y = Cx² - 3x

Solving Math Problems (Part 9)

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

Section 1.2, Problem 27

Solve the initial value problem:

L(dI/dt) + RI = 0 with I(0) = I₀

This is a problem out of electric circuit analysis.  We separate the variables to get:

Integrating both sides gives:

ln | I | = (R / L) t + C

Raising e to the power of both sides gives:

Solving for the given initial condition, we find the C = I₀ and the solution is:

Section 1.2, Problem 28

Solve the initial value problem:

v(dv/dx) = g = const with v(x₀) = v₀

We can proceed directly to integrating both sides:

∫ v dv = ∫ g dx

½v² = x g + C

Using the given initial condition, we solve for C to find C = ½(v₀² - 2 x₀ g) and the solution is:

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

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