When you have eliminated the impossible, whatever remains, however improbable, must be the truth, Sherlock Holmes.
Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
It involves (e.g., $\frac{dy}{dx} = 3x +5y$):
Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:
These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.
If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.
The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:
Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}), meaning that it satisfies the initial condition y(x_{0}) = y_{0}.
This theorem ensures that under these conditions, the solution exists and is unique near x = x_{0}.
Definition. A homogenous first-order ODE is a differential equation where the right-hand side can be expressed as a function of the ratio $\frac{y}{x}$. The general form is $y' = \frac{dy}{dx} = F(\frac{y}{x})$.
This form implies that the equation is scale-invariant, meaning if both x and y are scaled by the same non-zero constant, the form of the equation remains unchanged. This makes homogeneous ODEs well-suited for certain substitution methods that simplify their solutions.
Let us rescale both x and y by the same factor a, i.e., x = ax_{1}, y = ay_{1}, where a is a non-zero constant (a ≠ 0).
Calculate the derivatives: dx = adx_{1}, dy = ady_{1} ⇒ $\frac{dy}{dx} = F(\frac{y}{x}) ↭ \frac{ady_1}{adx_1} = \frac{dy_1}{dx_1} = F(\frac{y}{x}) = F(\frac{ay_1}{ax_1}) = F(\frac{y_1}{x_1}) ↭[\text{ the scaled version of the equation becomes:}] \frac{dy_1}{dx_1} = F(\frac{y_1}{x_1})$
This shows that the form of the equation remains the same after scaling, making the equation invariant under zoom. In simple terms, zooming in or out on the graph of the solution (scaling both axes equally) does not change the equation.
Consider the equation $y’ = \frac{x^2y}{x^3+y^3}$ ↭[To check if it is homogeneous, we divide both numerator and denominator by x^{3}] $y’ = \frac{\frac{y}{x}}{1+(\frac{y}{x})^3}$ The right-hand side of the equation is now a function of $\frac{y}{x}$, confirming that the ODE is homogeneous.
Consider the differential equation $x\frac{dy}{dx} = y + 2\sqrt{xy}$↭[To check if it is homogeneous, we divide both numerator and denominator by x] $\frac{dy}{dx} = \frac{y}{x} + \frac{2\sqrt{xy}}{x} ↭[\text{Assume that x is positive, then~} \sqrt{x^2}=|x| = x] \frac{dy}{dx} = \frac{y}{x} + \frac{2\sqrt{xy}}{\sqrt{x^2}} ↭ \frac{dy}{dx} = \frac{y}{x} + 2\sqrt{\frac{y}{x}}$. The right-hand side of the equation is now a function of $\frac{y}{x}$, confirming that the ODE is homogeneous.
Consider a second equation $xy’ = \sqrt{x^2 + y^2}↭ [\text{Divide both sides by x gives}] y’ = \sqrt{1+(\frac{y}{x})^2}$. Again, the right-hand side depends solely on $\frac{y}{x}$, confirming homogeneity.
Substitution. To solve an ODE of the form y’ = F($\frac{y}{x}$), the substitution z = $\frac{y}{x}$ is used, where z is a new variable representing the ratio $\frac{y}{x}$. This leverages the homogeneity to simplify the equation.
Differentiation. Consider y’ = F(y/x). If we define z = y/x, and use an inverse substitution y = zx ⇒[By the chain rule] y’ = z’x + z where $z’ = \frac{dz}{dx}$.
Substituting back into the original ODE. y’ = F(y/x), we get: z’x + z = F(z) ↭[This can be rearranged as:] $x\frac{dz}{dx} = F(z)-z$. This equation is separable.
Rewrite the equation to separate z and x: $\frac{dz}{F(z)-z} = \frac{dx}{x}$.
Integrate both sides. $\int \frac{dz}{F(z)-z} = \int \frac{dx}{x}$. Solve the integrals to find z as a function of x, considering any necessary techniques (e.g., partial fractions, substitution).
Back-Substitution. Once z(x) is found, substitute back z = $\frac{y}{x}$ to compute y(x).
Apply initial conditions if provided, to obtain the particular solution.
Rewrite the Equation. $(xy+y^2+x^2)dx -x^2dy = 0↭ x^2dy = (xy +y^2+x^2)dx$. Divide both sides by x^{2}dx, assuming x ≠ 0: $\frac{dy}{dx} = \frac{y}{x} + (\frac{y}{x})^2 + 1$ This confirms that the ODE is homogeneous.
Substitution, z = $\frac{y}{x}$. This substitution leverages the scale-invariance property of homogeneous first-order ODEs.
Differentiation. y = zx ⇒[By the chain rule] y’ = z’x + z where $z’ = \frac{dz}{dx}$.
Substituting back into the original ODE. y’ = F(y/x), we get: $z’x + z = z + z^2 +1$ ↭[This can be rearranged and simplified as:] $x\frac{dz}{dx} = z^2 +1↭ $. This equation is separable.
Separate variables: $\frac{dz}{z^2+1} = \frac{dx}{x}$.
Integrate both sides. $\int \frac{dz}{z^2+1} = \int \frac{dx}{x}↭ tan^{-1}(z) = ln|x| + C ↭ tan(tan^{-1}(z)) = tan(ln|x| + C) ↭ z = tan(ln|x| + C)$.
Back-Substitution. Once z(x) is found, substitute back z = $\frac{y}{x}$ to compute y(x). $\frac{y}{x} = tan(ln|x| + C)$ ⇒Therefore, the general solution to the differential equation is: $y = x·tan(ln|x| + C)$.
A lighthouse tracks a hostile boat (this is not just a boat; it’s a floating fortress of illicit activity, where trust is scarce and betrayal lurks just beneath the surface) moving at a constant angle 45° to the light beam. The lighthouse adjusts its beam to follow the boat. We aim to determine the boat’s path y = y(x).
Modeling the Boat’s Path
Let α be the angle between the boat’s position vector and the x-axis. From the geometry of the problem, we have: tan(α) = y/x (Refer to Figure D for a visual representation and aid in understanding it).
Since the boat moves at an angle of 45° relative to the light beam, the slope of the boat’s path is: y’ = $\frac{dy}{dx}$ = tan(α + 45°). This is because the boat’s direction is offset by 45° from the line connecting it to the lighthouse.
Using the angle addition formula for tangent tan(A + B) = $\frac{tan(A)+tan(B)}{1-tan(A)·tan(B)}$ and tan(45°) = 1, we have tan(α + 45°) = $\frac{tan(α)+tan(45)}{1-tan(α)·tan(45)} = \frac{tan(α)+1}{1-tan(α)} = [\text{Substitute tan(α)=y/x}] \frac{\frac{y}{x}+1}{1-\frac{y}{x}}$ This equation is homogeneous because the right-hand side is a function of $\frac{y}{x}$.
Substitute Back into the ODE: Let z = y/x, so y = zx. Differentiate both sides with respect to to x: y’ = z’x + z.
Substituting into the ODE, we have: $z’x + z = \frac{z+1}{1-z} ↭[\text{Rearrange the equation}] x\frac{dz}{dx} = \frac{z+1}{1-z}-z = \frac{z+1}{1-z}-\frac{z(1-z)}{1-z}=\frac{z+1-z+z^2}{1-z} = \frac{1+z^2}{1-z} ↭ x\frac{dz}{dx} = \frac{1+z^2}{1-z}$
Separation of variables. $\frac{1-z}{1+z^2}dz = \frac{dx}{x}$
Integrate both sides. $\int \frac{1-z}{1+z^2}dz = \int \frac{dx}{x}$
The left-hand side can be split into two integrals:
$\int \frac{1-z}{1+z^2}dz = \int \frac{1}{1+z^2}dz + \int \frac{-z}{1+z^2}dz = tan^{-1}(z)-\frac{1}{2}ln(1+z^2)$ because $\int \frac{z}{1+z^2}dz = \frac{1}{2}ln(1+z^2)$
Combined Integral Equation: $tan^{-1}(z)-\frac{1}{2}ln(1+z^2) = ln|x| + C ↭ tan^{-1}(z) = \frac{1}{2}ln(1+z^2) + ln|x| + C ↭tan^{-1}(z) = ln(\sqrt{1+z^2}) + ln|x| + C$.
Back-Substitution, z = y/x, $tan^{-1}(\frac{y}{x}) = \sqrt{ln(1+z^2)}+ln|x| + C = ln(\sqrt{1+(\frac{y}{x})^2})+ln(x) + C = ln(\frac{\sqrt{x^2+y^2}}{|x|})+ln|x| + C =[\text{Recall that ln(a/b) = ln(a)-ln(b)}] tan^{-1}(\frac{y}{x}) = ln(\sqrt{(x^2+y^2)})-ln|x|+ln|x|+C ↭[\text{The ln|x| terms cancel out}] tan^{-1}(\frac{y}{x}) = ln(\sqrt{x^2+y^2}) + c$
Converting to Polar Coordinates. In polar coordinates, r is the radial distance and θ is the angle, with $r=\sqrt{x^2+y^2}, θ = tan^{-1}(\frac{y}{x})$, so our equation becomes $θ = ln(r)+C$⇒[Exponentiate both sides] $e^θ = r·e^C ⇒[\text{Solve for r}] r = C_1e^θ$, where C_{1} is another constant.
Thus, the boat’s path is described by the equation $r = C_1e^θ$, which is the equation of an exponential spiral. It is a type of spiral curve that grows outward as it revolves around a central point. The defining characteristic is that the distance from the center increases exponentially as the angle increases.
An exponential spiral, also known as a logarithmic spiral, is a similar spiral curve which often appears in nature, such as in the shells of mollusks, hurricanes, and galaxies. In this context, it models the path of a boat moving away from a lighthouse while maintaining a constant angle with the beam.