Now I Am Become Death, the Destroyer of Worlds, Oppenheimer.
Do, or do not. There is no try, Yoda.
Definition. A differential equation is an equation that involves one or more functions and their derivatives. It relates the function itself (dependent variable), its derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
Key Components (e.g., $\frac{dy}{dx} = 3x +5y$):
Separable differential equations are a special class of differential equations that can be manipulated algebraically to separate the variables, allowing us to integrate and find solutions.
A separable differential equation is any differential equation that can be written in the form $\frac{dy}{dx}=f(x)g(y)$. This form allows us to separate the variables x and y on opposite sides of the equation, facilitating integration.
Definition. A first-order differential equation is called linear if it can be written in the form $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) and Q(x) are continuous functions of x over some interval I. This form only involves the first derivative of y with respect to x, hence it is called a first-order differential equation.
These equations fit the standard linear form $\frac{dy}{dx} +P(x)y = Q(x)$:
These equations are not linear due to the presence of non-linear terms in y or y′:
The integrating factor method is a useful technique for solving first-order linear ordinary differential equations (ODEs). This method involves multiplying the entire equation by a carefully chosen function, called the integrating factor. The purpose of this function, denoted μ(x), is to transform the left side of the equation into the derivative of a product, which simplifies integration.
First, we multiply our equation for the integrating factor: uy’ + P(x)u(x)y = uQ(x)
We aim to get (uy)’ = uy’ + P(x)uy and this only works if u’y = P(x)uy ↭ u’ = P(x)u ⇒[Separate variables] $\frac{u’}{u} = P ⇒[\text{Taking integrals}] \int \frac{du}{u} = \int P(x)dx ↭ lnu = \int P(x)dx ⇒[\text{Taking exponentials}] u = e^{\int P(x)dx}$
We are not going to use any arbitrary constant since only one integrating factor u is required. The integrating factor, μ(x), is defined as $μ(x) = e^{\int P(x)dx}$
The solution to the differential equation is obtained by multiplying both sides of the equation by the integrating factor $μ(\frac{dy}{dx} +P(x)y) = μ·Q(x)$, which becomes $\frac{d}{dx}(μ·y)=μ·Q(x)$ (💡Recognize the left side as the derivative of a product.) and then integrating both sides with respect to x (This transformation allows us to integrate both sides easily and find the solution), $μ·y = \int μ·Q(x)dx$, and now the ODE can be solved.
The equation is already in standard form: $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) = 1 and Q(x) = 5x^{2}.
Second, we compute the integrating factor, μ(x). It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int dx} = e^{x}$
Third, multiply both sides of the equation by the integrating factors: $e^xy’ + e^xy = e^x5x^2$ ↭[Recall $\frac{d}{dx}(μ·y)=μ·Q(x)$, the left-hand side is the derivative of the product μ(x)=e^{x} and y] $\frac{d}{dx}(e^x·y) = e^x5x^2$
Next, we can solve the differential equation integrating with respect to x, $\int \frac{d}{dx}(e^x·y)dx = \int e^x5x^2dx ↭ e^x·y = \int e^x5x^2dx$
$\int e^x5x^2dx = $[Integration by parts ∫udv=uv−∫vdu, u = 5x^{2} ⇒ du = 10xdx; dv = e^{x}dx ⇒ v = e^{x}] = $5x^2e^x -\int 10xe^xdx$ [Integration by parts ∫udv=uv−∫vdu, u = 10x ⇒ du = 10dx; dv = e^{x}dx ⇒ v = e^{x}] $5x^2e^x -(10xe^x-\int 10e^xdx) = 5x^2e^x-10xe^x +10e^x + C$
Therefore, $e^x·y = \int e^x5x^2dx = 5x^2e^x-10xe^x +10e^x + C ⇒[\text{Divide both sides by eˣ to solve for y}] y = 5x^2-10x+10 + Ce^{-x}$
The equation is already in standard form: $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) = 2 and Q(x) = x^{2}.
Second, we compute the integrating factor, μ(x). It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int 2dx} = e^{2x}$
Third, we write the differential equation in the form $\frac{d}{dx}(μ·y)=μ·Q(x)$ by multiplying both sides of the differential equation by the integrating factor, $e^{2x}(\frac{dy}{dx} + 2y) = e^{2x}·x^2 ↭ \frac{d}{dx}(e^{2x}·y) = e^{2x}·x^2$.
Next, we can solve the differential equation by integrating with respect to x, $e^{2x}·y = \int e^{2x}·x^2 dx$ =[Recall Integration by parts formula $\int udv = uv - \int vdu$, u = x^{2} (so du = 2xdx) dv = $e^{2x}dx$, so v = $\frac{1}{2}e^{2x}$] $x^2\frac{1}{2}e^{2x}- \int \frac{1}{2}e^{2x}2xdx = \frac{1}{2}x^2e^{2x}- \int e^{2x}xdx$
$\int e^{2x}xdx$ =[Apply integration by parts again, u = x, dv = $e^{2x}dx, du = dx, v = \frac{1}{2}e^{2x}$] $\frac{1}{2}xe^{2x} -\int \frac{1}{2}e^{2x}dx = \frac{1}{2}xe^{2x} -\frac{1}{4}e^{2x} + C.$ Therefore, $e^{2x}·y = \frac{1}{2}x^2e^{2x}- \int e^{2x}xdx = \frac{1}{2}x^2e^{2x} -\frac{1}{2}xe^{2x} +\frac{1}{4}e^{2x} - C = e^{2x}(\frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{4}) - C$
Finally, we can solve for y: $y = \frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{4} - C·e^{-2x}$ where C is a constant determined by initial conditions, if provided.
This differential equation models processes like Newton’s Law of Cooling, where T is the temperature of an object over time, k is a positive constant, and T_{e} is the surrounding environment’s temperature.
If an initial condition T(0) = T_{0} is provided, we can determine C: $T_0 = e^{-0·t}\int_{0}^{0} kT_e(t)e^{kt}dt + C = C$. Thus, C = T_{0} and $T(t) = e^{-kt}\int_{0}^{t} kT_e(t_1)e^{kt_1}dt_1 + T_0e^{-kt}$.
If T_{e}(t) is bounded and k > 0, the exponential term $T_0e^{-kt}$ approaches zero as t → ∞, so $e^{-kt}\int_{0}^{t} kT_e(t_1)e^{kt_1}dt_1$ is the steady-state solution.
For constant external temperature T_{e}(t), T_{e}(t) = T_{e}, $T = e^{-kt}\int kT_e(t)e^{kt}dt + Ce^{-kt} = e^{-kt}kT_e\int e^{kt}dt + Ce^{-kt} = e^{-kt}kT_e·\frac{e^{kt}}{k} + Ce^{-kt} = T_e + Ce^{-kt}=$ [Using the initial condition T(0) = T_{0}] T(0) = T_{0} = T_{e} + C⇒ C = T_{0} - T_{e}⇒ The solution to the differential equation modelling Newton's Law of Cooling for constant external temperature T_{e} and initial condition T(0) = T_{0} is $T = T_e + (T_0-T_e)e^{-kt}$.
This solution describes the temperature T(t) approaching T_{e} as t → ∞, with the rate of approach determined by k.