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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}.
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 can convert 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:
Substitutions are particularly useful when
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. Then, $y = v·x^2 ⇒[\text{Differentiate using the product rule}] \frac{dy}{dx}=v·2x + x^2\frac{dv}{dx}$
Substitute Back into the Original Differential Equation: $v·2x + x^2\frac{dv}{dx} = \frac{2(vx^2)}{x}+cos(v)↭ v·2x + x^2\frac{dv}{dx} = 2vx + cos(v)↭ x^2\frac{dv}{dx} = cos(v)↭[\text{Separate variables}] \frac{dv}{cos(v)} = \frac{dx}{x^2} ↭[\text{Integrate both sides}] \int \frac{dv}{cos(v)} = \int x^{-2}dx ↭ \int sec(v)dv = \int \frac{x^2}{dx} ↭[😊] ln|sec(v)+tan(v)| = -\frac{1}{x}+C ↭[\text{Substitute back v =}\frac{y}{x^2}] ln|sec(\frac{y}{x^2})+tan(\frac{y}{x^2})| = -\frac{1}{x}+C$
😊 We multiply the integrand by (sec(v) + tan(v))/(sec(v) + tan(v)), which is equal to 1: ∫ sec(v) dv = ∫ sec(v) * (sec(v) + tan(v))/(sec(v) + tan(v)) dv = ∫ (sec²(v) + sec(v)tan(v))/(sec(v) + tan(v)) dv. Now, notice that the numerator is the derivative of the denominator: d/dv (sec(v) + tan(v)) = sec(v)tan(v) + sec²(v). Therefore, we have an integral of the form ∫ f’(v)/f(v) dv, which is equal to ln|f(v)| + C: ∫ (sec²(v) + sec(v)tan(v))/(sec(v) + tan(v)) dv = ln|sec(v) + tan(v)| + C
This equation is challenging to solve directly due to the presence of the squared term involving both x and y. To simplify it, we can use a substitution that will reduce the equation to a separable form.
The expression x−y appears inside the squared term. This suggests that substituting v = x −y may simplify the equation.
Substitution: Let v = x-y. Then, $\frac{dv}{dx} = 1 -\frac{dy}{dx}⇒ \frac{dy}{dx}= 1 -\frac{dv}{dx}$
Substitute Back into the Original Differential Equation: $1 -\frac{dv}{dx} = (v+5)^2 ↭[\text{Rearrange the Equation}] \frac{dv}{dx} = 1-(v+5)^2$.
This is already a separable differential equation: $\frac{dv}{1-(v+5)^2} = dx$. Simplify the denominator: $1-(v+5)^2 = 1 -v^2-10v -25 = -v^2-10v-24 = -(v^2+10v+24) = -(v+6)(v+4)$.
Perform Partial Fraction Decomposition: $\frac{dv}{1-(v+5)^2} = dx ↭ \frac{-dv}{(v+6)(v+4)} = dx$
Set up the partial fractions: $\frac{1}{(v+6)(v+4)} = \frac{A}{(v+4)} + \frac{B}{(v+6)}$
Multiply both sides by (v+4)(v+6): 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 ⇒$B = \frac{-1}{2}, A = -B = \frac{1}{2}$,
$(\frac{\frac{-1}{2}}{v+6} + \frac{\frac{1}{2}}{v+4})dv = -dx ↭[\text{Multiply by 2}] (\frac{-1}{v+6}+ \frac{1}{v+4})dv = -2dx ↭[\text{Integrate both sides}] \int (\frac{-1}{v+6}+ \frac{1}{v+4})dv = \int -2dx ↭ -ln|v+6| +ln|v+4| = -2x + c_1↭ ln|\frac{v+4}{v+6}| = -2x + c_1 ↭[\text{Exponentiate both sides}] |\frac{v+4}{v+6}| = e^{-2x+c_1} ↭ \frac{v+4}{v+6} = c_2e^{-2x}↭ v+4 = c_2e^{-2x}(v+6) ↭ v+4 = c_2e^{-2x}v +6c_2e^{-2x}$
Next, subtract ce^{-2x}v: $v - c_2e^{-2x}v = 6e^{-2x} -4↭ v(1-c_2e^{-2x}) = 6c_2e^{-2x} -4$
$[\text{Solve for v}] v = \frac{6c_2e^{-2x} -4}{1-c_2e^{-2x}} ↭[\text{Back-Substituting}] x -y = \frac{6c_2e^{-2x} -4}{1-c_2e^{-2x}} ↭y = x + \frac{4 - 6c_2e^{-2x}}{1-c_2e^{-2x}} ↭ y = x + \frac{4 - 6ce^{-2x}}{1-ce^{-2x}}$
Accounting for Singular Solutions. I have assumed that (v+6)(v+4) ≠ 0:
The complete set of solutions to the differential equation $\frac{dy}{dx} = (x -y +5)^2$ includes:
$\frac{dy}{dx} = \frac{d}{dx}(x+6) = 1. (x-y+5)^2 = (x-(x+6)+5)^2 = (-1)^2 = 1$
Consider a physical system where the internal temperature T of an object changes over time due to heat exchange with its environment. The environment has a constant external temperature M. The rate at which the internal temperature changes is given by the differential equation: $\frac{dT}{dt} = k(M^4-T^4)$ where:
This model is relevant in situations involving large temperature differences, such as radiative heat transfer where the heat exchange is proportional to the difference of the fourth powers of the temperatures.
Our goal is to simplify the given differential equation by rescaling the temperature variable to make it dimensionless, thereby removing the constant M from the equation and simplifying the analysis.
Step 1: Define a Dimensionless Temperature Variable. We aim to eliminate the units of temperature from the equation by introducing a dimensionless variable. Let’s define: $T_1 = \frac{T}{M}$.
This rescaling means that T_{1} now represents the ratio of the internal temperature to the external temperature. Since both T and M have units of temperature, T_{1} is dimensionless.
When T = M, T_{1} = 1, meaning the internal and external temperatures are equal. T_{1} > 1 indicates the internal temperature is higher than the external temperature. T_{1} < 1 indicates the internal temperature is lower than the external temperature.
Step 2. Express T in terms of T_{1}: T = MT_{1}.
Step 3. Substitute T = MT_{1} into the original differential equation. $\frac{dT}{dt} = \frac{d(MT_1)}{dt} = M\frac{dT_1}{dt} = k(M^4-M^4T_1^4) = kM^4(1-T_1^4)↭ M\frac{dT_1}{dt} = kM^4(1-T_1^4) ↭ [\text{Dividing by M:}] \frac{dT_1}{dt} = kM^3(1-T_1^4)$
Step 4. Introduce a new constant to simplify further: $k_1 = kM^3$, which simplifies the equation: $\frac{dT_1}{dt} = k_1(1-T_1^4)$ where:
By making this substitution, we’ve simplified the original equation by removing M, leaving a more manageable equation to solve.
For simplicity, let’s denote T_{1} as T and k_{1} as k, so the equation becomes: $\frac{dT}{dt} = k(1-T^4)$. Our task is to solve this differential equation for T(t).
Separate variables: $\frac{dT}{1-T^4} = kdt$.
Integrate both sides: $\int \frac{dT}{1-T^4} = \int kdt$
Solve the integral. The integral on the left can be challenging due to the quartic term T^{4}. However, it can be simplified using partial fractions.
First, factor the denominator: $1-T^4 = (1-T^2)(1+T^2) = (1-T)(1+T)(1+T^2).$
The integral becomes: $\int \frac{dT}{1-T^4} = \int \frac{dT}{(1-T)(1+T)(1+T^2)} = \int kdt$
$\frac{1}{(1-T)(1+T)(1+T^2)} = \frac{A}{1-T}+ \frac{B}{1+T} + \frac{CT+D}{1+T^2}$
Multiply both sides by $(1-T)(1+T)(1+T^2): 1 = A(1+T)(1+T^2) + B(1-T)(1+T^2)+(CT+D)(1-T)(1+T)$
Expand and Simplify Each Term: $A(1+T)(1+T^2) = A[1·(1+T^2)+T·(1+T^2)] = A(1+T^2+T+T^3) = A(1+T+T^2+T^3); B(1-T)(1+T^2) = B[1·(1+T^2)-T(1+T^2)] = B(1 + T^2 -T - T^3) = B(1-T+T^2-T^3), (CT+D)(1-T)(1+T) = (CT+D)(1-T^2) = (CT)(1-T^2) + D(1-T^2) = (CT-CT^3)+(D-DT^2)$
$1 = A(1+T+T^2+T^3) + B(1-T+T^2-T^3) + (CT-CT^3)+(D-DT^2)$
Group Like Terms: (constant, T, T^{2}, T^{3}) $\begin{cases} A + B + D = 1 (i) \\ A -B + C = 0 (ii)\\ A +B -D = 0 (iii) \\ A -B -C = 0 (iv)\end{cases}$
Subtract equation (iv) from equation (ii): A -B + C -(A -B -C) = A - B + C -A + B + C = 0 ↭ 2C = 0 ↭ C = 0.
Substitute C = 0 into equation (2): A -B + C = 0 ⇒ A = B ⇒[Substitute A = B into (i) and (iii)] 2A + D = 1, 2A -D = 0 ⇒[Solve for A and D] $4A = 1 ↭ A = B = \frac{1}{4}$ ⇒[Now calculate D using 2A -D = 0] $D = \frac{1}{2}$
$\frac{1}{(1-T)(1+T)(1+T^2)} = \frac{A}{1-T}+ \frac{B}{1+T} + \frac{CT+D}{1+T^2} = \frac{1}{4}\frac{1}{1-T}+\frac{1}{4}\frac{1}{1+T}+\frac{\frac{1}{2}}{1+T^2}$
$\int (\frac{1}{4}\frac{1}{1-T}+\frac{1}{4}\frac{1}{1+T}+\frac{\frac{1}{2}}{1+T^2})dT = \int kdt ↭ -\frac{1}{4}ln|T-1|+\frac{1}{4}ln|1+T| + \frac{1}{2}arctan(T) = kt + C$.
Note: $\int \frac{1}{4}\frac{1}{1-T}dT =[u = 1-T, du = -dt] \frac{-1}{4}\int \frac{1}{u}du = \frac{-1}{4}ln|u| = \frac{-1}{4}ln|1-T|$
Final Solution. The implicit solution to the differential equation is: $-\frac{1}{4}ln|T-1|+\frac{1}{4}ln|1+T| + \frac{1}{2}arctan(T) = kt + C$ where C is the constant of integration.
Interpretation. The solution describes how the dimensionless temperature T evolves over time. Due to the transcendental nature of logarithmic and inverse trigonometric functions, solving for T explicitly in terms of t may not be feasible.
Bernoulli differential equations are a special class of nonlinear first-order differential equations of the form: y' = p(x)y + q(x)y^{n} where:
These equations are nonlinear due to the term y^{n} and can be challenging to solve directly. However, they can be transformed into linear differential equations using an appropriate substitution, making them solvable using standard methods.
Given the Bernoulli equation: y’ = p(x)y + q(x)y^{n}, our goal is to eliminate the nonlinear term y^{n} through substitution.
Let’s apply the Bernoulli substitution method to solve the differential equation: $y’ = \frac{y}{x}-y^2$.
Step 1: Rewrite the equation in standard form. $y’-\frac{y}{x} = -y^2$. We identify that this is a Bernoulli equation y’ + p(x)y = q(x)y^{n} with $p(x) = \frac{-1}{x}, q(x)=-1$, and n = 2. Since n = 2 ≠ 0, 1, we can proceed with the Bernoulli substitution method.
Step 2: Divide both sides by y^{2} (n = 2): $\frac{y’}{y^2} -\frac{1}{xy} = -1$ (🚀)
Step 3: Introduce the Substitution. We define a new variable $v = \frac{1}{y}$ (since v = y^{1-n} = y^{1-2} = y^{-1}). Compute the derivative $v’ = \frac{d}{dx}(\frac{1}{y}) = \frac{-1}{y^2}y’$. Solving for y: $y’ = -y^2v’ $
Step 4: Substitute Back into the Original Equation (🚀): $-v’ - \frac{v}{x} = -1$, and we rewrite it in standard form (multiply both sides by -1) $v’ + \frac{v}{x} = 1$
Step 5: Solve the linear equation:
Now, this is a linear first-order differential equation in v. The standard method for solving this is to find an integrating factor. The integrating factor is $μ(x)=e^{\int p(x)dx} = e^{\int \frac{1}{x}dx} = e^{ln(x)} = x$ (assuming x > 0 to avoid absolute values).
Multiply through by the integrating factor: v’x + v = x ↭[Recognize the left-hand side as a product derivative] So the equation becomes (xv)’ = x
Integrate both sides of the equation with respect to x: $xv = \int xdx = \frac{x^2}{2}+C⇒[\text{Solve for v}] v = \frac{1}{2}x+ \frac{C}{x} = \frac{x^2+2C}{2x}$⇒ Now, we need to backtrack our substitution: $v = \frac{1}{y},\text{ so } y = \frac{2x}{x^2+c_1}$ where c_{1} = 2c (for simplicity), and this is a constant of integration determined by initial conditions, if provided.