(I) Introduction

The 1st-order D.E. we have solved in previous sections have been either separable or linear. But many are not. In this section, we illustrate substitution methods that sometimes can be used to transform a given D.E. into one that we already know how to solve.

\( \frac{dy}{dx} = f(x,y) \)

If it contains a conspicuous combination \( v = \alpha(x,y) \), we may consider using \( v \) as a new variable.

EXAMPLE 1

Solve the D.E.: \( \frac{dy}{dx} = (x + y + 3)^2 \)

Let the substitution be: \( v = x + y + 3 \)

Then \( y = v - x - 3 \) and \( \frac{dy}{dx} = \frac{dv}{dx} - 1 \). Substituting into the D.E.:

\( \frac{dv}{dx} - 1 = v^2 \implies \frac{dv}{dx} = 1 + v^2 \)

This is now separable: \( \int \frac{dv}{1+v^2} = \int dx \implies \arctan v = x + C \)

General solution: \( y = \tan(x + C) - x - 3 \)

Remark: Any D.E. of the form \( \frac{dy}{dx} = F(ax + by + c) \) can be transformed into a separable equation by use of substitutions \( v = ax + by + c \).

(II) Homogeneous Equations

Standard form of a first order homogeneous equation:

$$ \frac{dy}{dx} = F\left(\frac{y}{x}\right) $$

Make substitution:

$$ v = \frac{y}{x} \implies y = vx $$

Applying the product rule: \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

Remark: \( Ax^m y^n \frac{dy}{dx} = Bx^p y^q + Cx^r y^s \) is "homogeneous" if each term has the same total degree: \( m+n = p+q = r+s \).

EXAMPLE 2

Solve: \( (x - y)y' = x + y \)

Dividing by \(x\): \( (1 - \frac{y}{x})y' = 1 + \frac{y}{x} \)

Let \( v = y/x \), and \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

\( (1 - v)(v + x\frac{dv}{dx}) = 1 + v \implies x\frac{dv}{dx} = \frac{1 + v^2}{1 - v} \)

\( \int \frac{1-v}{1+v^2} dv = \int \frac{dx}{x} \implies \arctan v - \frac{1}{2}\ln(1+v^2) = \ln|x| + C \)

General Solution: \( \arctan \frac{y}{x} = \frac{1}{2}\ln(x^2 + y^2) + C \)

(III) Bernoulli Equations

A 1st-order D.E. of the form:

$$\frac{dy}{dx} + P(x)y = Q(x)y^n $$

is called a Bernoulli equation. When \( n=0 \) or \( n=1 \), the equation is linear. For other values of \( n \), we use a substitution to transform it into a linear equation.

Detailed Derivation of the General Formula

1. Division by \( y^n \): Multiply the entire equation by \( y^{-n} \):

$$ y^{-n} \frac{dy}{dx} + P(x)y^{1-n} = Q(x) $$

2. Substitution: Let \( v \) be the new dependent variable:

$$ v = y^{1-n} $$

3. Differentiation: \( \frac{dv}{dx} = (1-n)y^{-n} \frac{dy}{dx} \implies y^{-n} \frac{dy}{dx} = \frac{1}{1-n} \frac{dv}{dx} \)

4. Transformation: \( \frac{1}{1-n} \frac{dv}{dx} + P(x)v = Q(x) \)

5. Standard Linear Form: Multiply by \( (1-n) \) to obtain a linear 1st-order D.E.:

$$\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$$

EXAMPLE 3

Solve: \( y^2 y' + 2xy^3 = 6x \)

Let \( v = y^3 \), so \( \frac{dv}{dx} = 3y^2 \frac{dy}{dx} \):

\( \frac{1}{3}\frac{dv}{dx} + 2xv = 6x \implies \frac{dv}{dx} + 6xv = 18x \)

Integrating Factor: \( \rho(x) = e^{\int 6x dx} = e^{3x^2} \)

\( D_x(v e^{3x^2}) = 18x e^{3x^2} \implies v e^{3x^2} = 3e^{3x^2} + C \)

General solution: \( y = \sqrt[3]{3 + Ce^{-3x^2}} \)

(IV) Exact Differential Equations

The implicit general solution of a 1st-order D.E. is \( F[x, y(x)] = C \). Taking the derivative with respect to \( x \) yields \( F_x + F_y \frac{dy}{dx} = 0 \). This corresponds to a 1st-order D.E. of the form: \( M(x, y) dx + N(x, y) dy = 0 \)

Definition

The differential form \( M(x, y) dx + N(x, y) dy = 0 \) is called an exact differential equation if there exists \( F(x, y) \) such that: \( F_x = M \) and \( F_y = N \). \( F(x, y) = C \) is the general solution.

Theorem 1 (Criterion for Exactness)

Suppose that \( M(x, y) \) and \( N(x, y) \) are continuous and have continuous 1st-order partial derivatives. Then the D.E. is exact iff:

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$
at each point of the rectangle \( R \).

EXAMPLE 4: Comparison

1. \( y^3 dx + 3xy^2 dy = 0 \implies \frac{\partial M}{\partial y} = 3y^2, \quad \frac{\partial N}{\partial x} = 3y^2 \implies \text{Exact!} \)

2. \( y dx + 3x dy = 0 \implies \frac{\partial M}{\partial y} = 1, \quad \frac{\partial N}{\partial x} = 3 \implies \text{Not exact!} \)

Note: Both equations are equivalent and have the same solution: \( y^3 x = C \).

Remark: whether a D.E. is exact or not is related to the precise form \( M dx + N dy = 0 \) in which it's written.

EXAMPLE 5

Solve: \( e^y + y \cos x + (x e^y + \sin x + 1) y' = 0 \)

1. Verify Exactness: \( \frac{\partial M}{\partial y} = e^y + \cos x, \quad \frac{\partial N}{\partial x} = e^y + \cos x \implies \text{Exact!} \)

2. Find \( F \): \( F(x, y) = \int (e^y + y \cos x) dx + g(y) = x e^y + y \sin x + g(y) \)

3. Solve for \( g(y) \): \( F_y = x e^y + \sin x + g'(y) = N = x e^y + \sin x + 1 \implies g'(y) = 1 \implies g(y) = y \)

General solution: \( x e^y + y \sin x + y = C \)

(V) Reducible 2nd order D.E.

1. Dependent variable \( y \) missing: \( F(x, y', y'') = 0 \)

Let \( p = y' \), then \( y'' = \frac{dp}{dx} \).

EXAMPLE 6

Solve: \( xy'' + 2y' = 6x \)

Let \( p = y' \): \( x\frac{dp}{dx} + 2p = 6x \implies \frac{dp}{dx} + \frac{2}{x}p = 6 \). Integrating Factor: \( \rho(x) = x^2 \)

\( D_x(x^2 p) = 6x^2 \implies x^2 p = 2x^3 + C_1 \implies p = \frac{dy}{dx}=2x + C_1 x^{-2} \)

General solution: \( y(x) = x^2 - \frac{C_1}{x} + C_2 \)

2. Independent variable \( x \) missing: \( F(y, y', y'') = 0 \)

Let \( p = y' \), then \( y'' = p \frac{dp}{dy} \).

EXAMPLE 7

Solve: \( yy'' + (y')^2 = 0 \)

Let \( p = y' \): \( y p \frac{dp}{dy} + p^2 = 0 \implies \int \frac{dp}{p} = \int -\frac{dy}{y} \implies py = C_1 \)

\( y \frac{dy}{dx} = C_1 \implies \int y dy = \int C_1 dx \implies y^2 = Ax + B \)

Extra: Non-Homogeneous to Homogeneous

Extra Example

Solve the D.E.: \( \frac{dy}{dx} = \frac{2y - x + 7}{4x - 3y - 18} \).

1. Find translation constants \( h \) and \( k \) by setting constant terms to zero:
\( 2k - h + 7 = 0 \) and \( 4h - 3k - 18 = 0 \implies h = 3, \quad k = -2 \).
Substitutions: \( x = u + 3 \) and \( y = v - 2 \).

2. Homogeneous Substitution:
The equation becomes \( \frac{dv}{du} = \frac{2v - u}{4u - 3v} = \frac{-1 + 2(v/u)}{4 - 3(v/u)} \).
Let \( w = \frac{v}{u} \implies v = uw \implies \frac{dv}{du} = w + u\frac{dw}{du} \).
Substituting: \( w + u\frac{dw}{du} = \frac{-1 + 2w}{4 - 3w} \implies u\frac{dw}{du} = \frac{3w^2 - 2w - 1}{4 - 3w} \).

3. Integrate by Partial Fractions:
\( \int \frac{4 - 3w}{3w^2 - 2w - 1} \, dw = \int \frac{du}{u} \).
Using partial fraction decomposition:
\( \int \left( \frac{1}{4} \cdot \frac{1}{w-1} - \frac{15}{4} \cdot \frac{1}{3w+1} \right) \, dw = \int \frac{du}{u} \).
\( \frac{1}{4} \ln|w-1| - \frac{5}{4} \ln|3w+1| = \ln|u| + C_1 \).

4. Back-Substitution:
Combine logarithms: \( \ln \frac{|w-1|}{|3w+1|^5} = \ln|u|^4 + C_2 \implies \frac{w-1}{(3w+1)^5 u^4} = C \).
Substitute \( w = v/u \): \( \frac{v/u - 1}{(3v/u + 1)^5 u^4} = C \implies \frac{v - u}{(3v + u)^5} = C \).
Replace \( u, v \) with original variables \( u = x - 3, v = y + 2 \):
\( \frac{(y + 2) - (x - 3)}{(3(y + 2) + (x - 3))^5} = C \implies \frac{y - x + 5}{(3y + x + 3)^5} = C \).

General Solution: \( \frac{y - x + 5}{(3y + x + 3)^5} = C \).