As long as algebra is taught in school, there will be prayer in school, Cokie Roberts

Recall. Algebraic and differential equations.

An algebraic equation is a mathematical statement asserting that two algebraic expressions are equal.. These expressions are constructed using:

Dependent and independent variables. Variables are symbols representing unknown quantities. Variables can be:
Independent Variables: Variables or input that can be chosen (or change) freely.
Dependent Variables: Variables that depend on the independent variables.
Constants. Fixed numerical values that do not change, e.g., 2, π, 7.
Algebraic operations. Operations such as addition, subtraction, multiplication, division, exponentiation, powers, and root.

An algebraic equation typically has the following general form: Expression₁ = Expression₂, where the two expressions are set equal to each other.

Examples: y = 2x + 4 (simple linear equation, represents a straight line with a slope of 2 and a y-intercept of 4), ax² + bx + c = 0 (quadratic equation, it represents a parabola), $ln(xy + t) = z·sin(\sqrt{x})$.

Definition. A differential equation is an equation that involves one or more functions and their derivatives. It relates the function itself (the dependent variable), its derivatives with respect to one or more independent variables (rates of change), and the independent variable(s) themselves, e.g., $\frac{dy}{dx} = 3x +5y, y' + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

If an algebraic equation is a “snapshot” of a relationship, a differential equation is the “instruction manual” for how that relationship moves and changes.

Key Components (e.g., $\frac{dy}{dx} = 3x +5y$):

Dependent variables: The function y (or y(x)) we are trying to find or solve for. Variable(s) (y) that depend on one or more other variables.
Independent variables: Variable(s) (x) upon which the dependent variables depend and we differentiate with respect to (often time or space).
Derivatives: Expressions like $\frac{dy}{dx}$ or y’ that represent the rate at which the dependent variable (y) change with respect to the independent variable (x).

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It answers two fundamental questions: Is there a solution? and Is it the only one? It states that if:

The function $f(x, y)$ (the right-hand side of the ODE when written as $y' = f(x, y)$) is continuous in some neighborhood around a specific point $(x_0, y_0)$ and
Its partial derivative with respect to y, denoted as $\frac{\partial f}{\partial y}$, is also continuous near $(x_0, y_0)$.

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it perfectly satisfies the initial condition $y(x_0) = y_0$.

This theorem is a mathematician’s safety net: it ensures that under these well-behaved conditions, a solution definitively exists and is absolutely unique near $x = x_0$. The physical system modeled by the ODE will only follow one distinct path.

Substitution Methods

First-order Substitution Methods

Differential equations are fundamental in modeling various physical phenomena in fields such as physics, engineering, biology, and economics. However, solving these equations can be quite challenging due to their complexity, especially when they involve nonlinear terms or non-standard forms. One effective technique to simplify and solve such equations is through substitution methods.

Substitution methods involve introducing new variables or rescaling existing ones to transform a complicated differential equation into a more manageable form. This technique acts as a mathematical disguise, converting a difficult equation into one that is separable, linear, or of a known type that can be solved using standard methods.

There are two main types of substitutions used in solving differential equations:

Direct substitution. In this approach, we define a new variable as a function of the original variables. This substitution simplifies the equation by replacing complex expressions with simpler ones, e.g., the integral $\int x\sqrt{1-x^2}dx$ is solved with the substitution $u = 1 -x^2$, which implies du = -2xdx. We can rewrite the integral entirely in terms of u, which drastically simplifies the integration process, $\int x\sqrt{1-x^2}dx = \frac{-1}{2}\int \sqrt{u}du$.
Inverse substitution: This method involves expressing the original variables as functions of a new variable, e.g., the integral $\int \sqrt{1-x^2}dx$ is solved by letting x = sin(u), dx = cos(u)du. The integral becomes $\int \sqrt{1-x^2}dx = \int \sqrt{1-sin^2(u)}·cos(u)du = \int cos^2(u)du$, which is easier to evaluate using trigonometric identities..

Substitutions are particularly useful when:

The differential equation involves nonlinear terms that are difficult or impossible to isolate directly.
We aim to eliminate constants to make the equation dimensionless, simplifying the analysis.
We need to reduce the equation to a known form that can be solved using standard methods.

Solving the Differential Equation $\frac{dy}{dx} = \frac{2y}{x}+cos(\frac{y}{x^2})$

We are tasked with solving a nonlinear differential equation. This equation is challenging to solve directly due to the presence of the term $\frac{y}{x^2}$ inside the cosine function and the nonlinear dependence on y. To simplify it, we can use a substitution that reduces the equation to a separable form.

We notice that y appears in the form $\frac{y}{x^2}$, this suggest the substitution: v = $\frac{y}{x^2}.$ By setting v equal to this expression, we aim to simplify the equation and reduce it to a separable form.

Step 1. Differentiate the substitution.

First, rewrite the substitution to isolate y: $y = v·x^2$.

Next, differentiate both sides with respect to x using the product rule: $\frac{dy}{dx}=v·2x + x^2\frac{dv}{dx}$

Step 2. Substitute Back into the Original Differential Equation.

$v·2x + x^2\frac{dv}{dx} = \frac{2(vx^2)}{x}+cos(v)↭[\text{Simplify the right side:}] v·2x + x^2\frac{dv}{dx} = 2vx + cos(v)$.

Subtract 2vx from both sides (notice how beautifully they cancel out!): $x^2\frac{dv}{dx} = cos(v)$

Step 3. Separate variables and integrate:

Now we have a standard separable equation. Move all v terms to the left and x terms to the right: $\frac{dv}{cos(v)} = \frac{dx}{x^2}$

Integrate both sides: $\int \frac{dv}{cos(v)} = \int x^{-2}dx ↭ \int sec(v)dv = \int \frac{x^2}{dx} ↭[😊] ln|sec(v)+tan(v)| = -\frac{1}{x}+C$

This equation is an implicit solution to the original differential equation. Due to the transcendental nature of the functions involved, solving for y explicitly in terms of x may not be possible.

😊 We multiply the integrand by $\frac{sec(v) + tan(v)}{sec(v) + tan(v)}$, which is equal to 1: $\int sec(v) dv = \int sec(v) \cdot \frac{sec(v) + tan(v)}{sec(v) + tan(v)} dv = \int \frac{sec²(v) + sec(v)tan(v)}{sec(v) + tan(v)}dv$. Now, notice that the numerator is the exact derivative of the denominator: $\frac{d}{dv}(sec(v) + tan(v)) = sec(v)tan(v) + sec²(v)$. Therefore, we have an integral of the form $\int \frac{f'(v)}{f(v)} dv$, which evaluates to $ln|f(v)| + C$, that is $\int \frac{sec²(v) + sec(v)tan(v)}{sec(v) + tan(v)} dv = ln|sec(v) + tan(v)| + C$.

Step 4. Back-substitute:

Finally, replace v with our original expression $v = \frac{y}{x^2}$: $\boxed{ln|sec(\frac{y}{x^2})+tan(\frac{y}{x^2})| = -\frac{1}{x}+C}$.

Note: This equation is an implicit solution to the original differential equation. Due to the transcendental nature of the trigonometric and logarithmic functions involved, solving for y explicitly in terms of x is not algebraically possible here.

Solve $\frac{dy}{dx} = (x -y +5)^2$

This equation is challenging to solve directly due to the presence of the squared term involving both x and y. Expanding the binomial would create a messy mixture of $x^2$, $y^2$, and $xy$ terms. Instead, we use substitution to simplify it.

The expression x−y appears neatly packaged inside the squared term. This suggests that substituting v = x −y may simplify the equation.

Step 1. Substitution: Let v = x-y. Differentiating both sides with respect to $x$ yields: $\frac{dv}{dx} = 1 -\frac{dy}{dx}⇒ \frac{dy}{dx}= 1 -\frac{dv}{dx}$

Step 2. Substitute Back into the Original Differential Equation: $1 -\frac{dv}{dx} = (v+5)^2 ↭[\text{Rearrange the equation to isolate the derivative}] \frac{dv}{dx} = 1-(v+5)^2$.

Step 3. Simplify and Separate Variables: This is already a separable differential equation: $\frac{dv}{1-(v+5)^2} = dx$. Simplify the denominator: $1-(v+5)^2 = 1 -v^2-10v -25 = -v^2-10v-24 = -(v^2+10v+24) = -(v+6)(v+4)$.

Step 4. Perform Partial Fraction Decomposition: $\frac{dv}{1-(v+5)^2} = dx ↭ \frac{-dv}{(v+6)(v+4)} = dx$

To integrate the left side, we set up partial fractions: $\frac{1}{(v+6)(v+4)} = \frac{A}{(v+4)} + \frac{B}{(v+6)}$

Multiply both sides by the equal denominator (v+4)(v+6), we get: 1 = A(v+6) +B(v+4) = Av + 6A +Bv +4B ↭ [Group like terms:] 1=(A+B)v+(6A+4B). Set up the system of equations by equating coefficients:

$\begin{cases} A + B = 0 \\\ 6A + 4B = 1 \end{cases}$

From equation (1): A = -B. Substitute A=−B into equation (2): 6(−B) +4B = 1 ⇒ −2B = 1 ⇒$B = \frac{-1}{2}, A = -B = \frac{1}{2}$,

Step 5. Integrate.

Substitute $A$ and $B$ back into the separated equation:

$$ \begin{aligned} \left( \frac{1/2}{v+4} - \frac{1/2}{v+6} \right)dv = -dx &\text{[Multiply by 2 to clear the fractions]} \\[2pt] &=\left( \frac{1}{v+4} - \frac{1}{v+6} \right)dv = -2dx \\[2pt] &\text{Integrate both sides}\\[2pt] &\ln|v+4| - \ln|v+6| = -2x + C_1 \\[2pt] &\iff ln\left|\frac{v+4}{v+6}\right| = -2x + C_1 \end{aligned} $$

Step 6.Solve for v:

Exponentiate both sides to remove the natural log:

$$ \begin{aligned} \left|\frac{v+4}{v+6}\right| = e^{-2x + C_1} = e^{C_1}e^{-2x} &\text{[Let } C_2 = \pm e^{C_1} \text{ to drop the absolute value bars]}\\[2pt] &\iff \frac{v+4}{v+6} = C_2e^{-2x} \\[2pt] &\text{Multiply by (v + 6) and distribute:}\\[2pt] &\iff v + 4 = C_2e^{-2x}(v+6) \\[2pt] &\text{Group the v terms on one side:}\\[2pt] &\iff v - C_2e^{-2x}v = 6C_2e^{-2x} - 4 \\[2pt] &\iff v(1 - C_2e^{-2x}) = 6C_2e^{-2x} - 4 \\[2pt] &[\text{Solve for v}] v = \frac{6C_2e^{-2x} - 4}{1 - C_2e^{-2x}} \end{aligned} $$

Step 7. Back-Substitute and Clean Up:

Recall that $v = x - y \implies y = x - v$: $y = x - \left( \frac{6C_2e^{-2x} - 4}{1 - C_2e^{-2x}} \right)$

Distribute the negative sign into the numerator: $y = x + \frac{4 - 6C_2e^{-2x}}{1 - C_2e^{-2x}}$ (Optional Algebraic Cleanup): While the above answer is correct, negative exponents can be visually messy. If we multiply the top and the bottom of the fraction by $\frac{-e^{2x}}{C_2}$, and define a new constant $c = \frac{-1}{C_2}$, the equation transforms into a much cleaner form: $\boxed{y = x + \frac{6 + 4ce^{2x}}{1 + ce^{2x}}}$

Accounting for Singular Solutions

During step 3, we divided both sides by $(v+6)(v+4)$. In doing to do so, we implicitly assumed that this expression does not equal zero. We must manually check what happens if it does equal zero!

Case 1: $v + 4 = 0 \implies x - y + 4 = 0 \implies y = x + 4$. Is this captured by our general solution? Yes! If we set our constant $c \to \infty$, the fraction simplifies to $\frac{4ce^{2x}}{ce^{2x}} = 4$, yielding $y = x + 4$.

Case 2: $v + 6 = 0 \implies x - y + 6 = 0 \implies$ y = x + 6. This is a singular solution. If you plug $c = 0$ into our final cleaned-up general formula, you actually get $y = x + 6$. However, depending on how you define your initial constant of integration, this solution often requires the constant to be zero or undefined, meaning it is easily lost during the separation of variables.

The complete set of solutions to the differential equation $\frac{dy}{dx} = (x - y + 5)^2$ is neatly summarized by:

The general solution family: $y = x + \frac{6 + 4ce^{2x}}{1 + ce^{2x}}$ (where $c$ is an arbitrary constant).
The bounding linear solutions: $y = x + 6$ and $y = x + 4$.

Verification of the Singular Solution: Let’s plug $y = x + 6$ back into the original ODE to prove it works. Left side: $\frac{dy}{dx} = \frac{d}{dx}(x+6) = 1$. Right side: $(x - y + 5)^2 = (x - (x+6) + 5)^2 = (-1)^2 = 1$. Since $1 = 1$, it is a perfectly valid solution!

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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