Solving the Second-Order Differential Equation y′′ y′ tan x y′^2

This article will guide you through the detailed process of solving the second-order differential equation y′′ y′ tan x y′^2. We will explore methods such as transformations, factorization, and integration to derive the general solution of this intriguing equation.

Understanding the Equation

The given differential equation is:

y′′ y′ tan x y′^2

Where y is a function of x, and y′ and y′′ represent the first and second derivatives of y with respect to x, respectively. Our goal is to express y in terms of x and a constant of integration.

Transformation to a First-Order Equation

We start by making a substitution to simplify the given differential equation. Let:

p y′

This implies:

y ∫p dx

Substituting this into the original equation, we get:

dy′/dx p tan x p^2

Simplifying, we have:

dy′/dx p^2 - p tan x

Further simplifying, we get a first-order equation in terms of p:

dy′/dx p(p - tan x)

Separation of Variables

The equation can now be written as:

dp/(p(p - tan x)) dx

We proceed by using partial fraction decomposition:

1/(p(p - tan x)) A/p B/(p - tan x)

Multiplying through by p(p - tan x), we get:

1 Ap - Atan x Bp - Btan x

Solving for A and B, we set up the system of equations:

A - B tan x 0

A B 1

Solving this system, we find:

A -1/tan x

B 1/tan x

Substituting back, we get:

1/(p(p - tan x)) -1/tan x (1/p 1/(p - tan x))

Integration and Solution

Thus, the equation becomes:

-1/tan x (1/p 1/(p - tan x)) dx

Integrating both sides, we obtain:

-1/tan x ln|p| - 1/tan x ln|p - tan x| x C

Combining the logarithms, we get:

ln|p - tan x|/|p| tan x (x C)

Exponentiating both sides, we find:

|p - tan x|/|p| e^(tan x (x C))

Removing the absolute values and rearranging, we get:

p - tan x p e^(tan x (x C))

Rearranging to solve for p, we have:

p (tan x)/(1 - e^(tan x (x C)))

Since y ∫p dx, the general solution for y is:

y ∫(tan x)/(1 - e^(tan x (x C))) dx C_1

Where C_1 is a constant of integration.

Alternative Solution

Another approach to solving this differential equation involves recognizing that y const. is a solution. This suggests that we might consider the equation:

y′ Vx

Where V is a function of x. Substituting this into the original equation, we get:

V′x V^2 tan x V^2

Simplifying, we find:

V′x V^2(1 - tan x)

Rearranging, we have:

V′/V^2 1/x(1 - tan x)

This is a Bernoulli equation. Let us make the substitution:

V 1/W

Substituting, we get:

W′/W^2 -1/x(1 - tan x)

Multiplying through by W, we obtain:

W′/W 1/xW -1/x(1 - tan x)

This is a linear first-order differential equation. The integrating factor is:

e^∫(-1/tan x)dx cos x

Multiplying both sides by the integrating factor, we get:

cos x(dW/dx 1/x W) -cos x

This simplifies to:

d(cos x W)/dx -cos x

Integrating both sides, we find:

cos x W -sin x C

Thus:

W C cos x - sin x

Since V 1/W, we have:

V 1/(C cos x - sin x)

And since y′ Vx, we get:

y ∫(x/(C cos x - sin x)) dx C_1

The general solution is:

y -ln|C cos x - sin x| C_2

Where C_1 and C_2 are constants of integration.