Differential Equations
Integral Calculus: Differential Equations
Foundations of Differential Equations
Order, Degree, and Formation of EquationsTopic 1
- Order: The order of a differential equation is the order of the highest-order derivative appearing in the equation. It is always a positive integer and is unconditionally defined for every DE.
- Degree: The degree of a differential equation is the power or exponent of the highest-order derivative, provided the equation can be expressed as a polynomial equation in terms of its derivatives. If the derivatives are embedded inside transcendental functions (such as $\sin(y')$, $\ln(y'')$, or $e^{y'}$), the degree of the equation is strictly undefined.
Determine the order and degree of the following differential equation: \[ y = x \frac{dy}{dx} + \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \]
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The highest-order derivative in the equation is the first derivative, $\frac{dy}{dx}$. Therefore, the order of the equation is 1. To find the degree, we must first remove the radical sign to express the equation as a polynomial in terms of its derivatives. Rearrange the terms: \[ y - x \frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \] Square both sides of the equation to eliminate the square root: \[ \left(y - x \frac{dy}{dx}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2 \] Expand the left side: \[ y^2 - 2xy \frac{dy}{dx} + x^2 \left(\frac{dy}{dx}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2 \] Gather all terms onto one side: \[ (x^2 - 1)\left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + (y^2 - 1) = 0 \] The equation is now expressed as a polynomial in terms of its derivatives. The exponent of the highest-order derivative ($\frac{dy}{dx}$) is 2. Therefore, the degree of the equation is 2. Final Answer: Order = 1, Degree = 2.
Solution Methodologies for First-Order Equations
Separation of Variables and SubstitutionsTopic 1
The most straightforward method for solving a first-order differential equation is the Variable Separable Form. If the equation can be factored such that the derivative $\frac{dy}{dx}$ splits into a product of a function of $x$ and a function of $y$, you can group all terms containing $y$ with the differential $dy$, and all terms containing $x$ with the differential $dx$: \[ \frac{dy}{dx} = f(x)g(y) \implies \frac{1}{g(y)} \, dy = f(x) \, dx \] Integrating both sides directly yields the general solution: $\int \frac{1}{g(y)} \, dy = \int f(x) \, dx + C$.
When a differential equation is not immediately separable, it can often be reduced to variable separable form using an algebraic substitution:- Linear Argument Forms: For equations of the type $\frac{dy}{dx} = f(ax + by + c)$, substitute $v = ax + by + c$. Differentiating this substitution with respect to $x$ yields $\frac{dv}{dx} = a + b\frac{dy}{dx}$, which transforms the original equation into a separable equation in terms of $v$ and $x$.
- Homogeneous Forms: A differential equation is homogeneous if it can be written in the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$. To solve this form, substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$. This transformation simplifies the equation into a separable form in terms of $v$ and $x$: \[ v + x\frac{dv}{dx} = F(v) \implies x\frac{dv}{dx} = F(v) - v \implies \frac{1}{F(v) - v} \, dv = \frac{1}{x} \, dx \]
Solve the following homogeneous differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.
Show solution
Divide the numerator and denominator by $x^2$ to express the right side as a function of the ratio $\frac{y}{x}$: \[ \frac{dy}{dx} = \frac{1 + \left(\frac{y}{x}\right)^2}{2\left(\frac{y}{x}\right)} \] Apply the standard homogeneous substitution: $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$. \[ v + x\frac{dv}{dx} = \frac{1 + v^2}{2v} \] Subtract $v$ from both sides to isolate the derivative term: \[ x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v} \] Separate the variables by moving all terms containing $v$ to the left side and all terms containing $x$ to the right side: \[ \frac{2v}{1 - v^2} \, dv = \frac{1}{x} \, dx \] Integrate both sides. Use substitution on the left side ($u = 1 - v^2 \implies du = -2v \, dv$): \[ -\ln|1 - v^2| = \ln|x| + C_1 \implies \ln|x| + \ln|1 - v^2| = -C_1 \] Combine the logarithmic terms: \[ \ln|x(1 - v^2)| = -C_1 \implies x(1 - v^2) = e^{-C_1} = C \] Substitute the original variable ratio $v = \frac{y}{x}$ back into the equation: \[ x\left(1 - \frac{y^2}{x^2}\right) = C \implies x\left(\frac{x^2 - y^2}{x^2}\right) = C \implies \frac{x^2 - y^2}{x} = C \implies x^2 - y^2 = Cx \] Final Answer: $x^2 - y^2 = Cx$.
Linear First-Order Equations, Bernoulli's Form, and ExactnessTopic 2
A first-order differential equation is linear if it can be written in the standard form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where $P(x)$ and $Q(x)$ are continuous functions of $x$. To solve this equation, multiply all terms by an Integrating Factor (IF), denoted by $\mu(x) = e^{\int P(x) \, dx}$. Multiplying by the integrating factor transforms the left side into the derivative of a product, allowing you to integrate directly: \[ \frac{d}{dx}\left[ y \cdot e^{\int P(x) \, dx} \right] = Q(x) \cdot e^{\int P(x) \, dx} \implies y \cdot \mu(x) = \int Q(x) \mu(x) \, dx + C \]
Two advanced variations of this first-order form are frequently tested in competitive exams:- Bernoulli's Equation: A non-linear equation containing a power of the dependent variable on the right side: $\frac{dy}{dx} + P(x)y = Q(x)y^n$. To linearize this equation, divide all terms by $y^n$ and substitute $v = y^{1-n}$. This transforms the expression into a standard linear equation in terms of $v$ and $x$.
- Exact Differential Equations: An equation expressed in differential form, $M(x,y) \, dx + N(x,y) \, dy = 0$, is exact if its components satisfy the partial derivative condition: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] The general solution is found by integrating $M$ with respect to $x$ (treating $y$ as a constant) and adding the integrals of the terms in $N$ that contain *only* $y$. If the exactness condition is not met, the equation can often be made exact by multiplying by a targeted integrating factor.
Find the general solution of the following linear differential equation: $\frac{dy}{dx} + \frac{2}{x}y = x^3 \quad (x > 0)$.
Show solution
- Step 1: Calculate the Integrating Factor (IF)
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2 \] - Step 2: Apply the linear solution formula
\[ y \cdot \mu(x) = \int Q(x)\mu(x) \, dx + C \] Substitute $\mu(x) = x^2$ and $Q(x) = x^3$ into the equation: \[ y \cdot x^2 = \int (x^3 \cdot x^2) \, dx + C = \int x^5 \, dx + C \] Integrate the right side using the standard power rule: \[ y \cdot x^2 = \frac{x^6}{6} + C \] Divide both sides by $x^2$ to isolate $y$ and find the general solution: \[ y = \frac{x^4}{6} + C x^{-2} \]
Higher-Order Systems & Advanced Solution Forms
Second-Order Linear Equations with Constant CoefficientsTopic 1
- Complementary Function ($y_c$): Found by setting the forcing function to zero ($f(x) = 0$) and solving the corresponding auxiliary quadratic equation $a m^2 + b m + c = 0$:
- Real and Distinct Roots ($m_1 \neq m_2$): $y_c = c_1 e^{m_1 x} + c_2 e^{m_2 x}$
- Real and Repeated Roots ($m_1 = m_2 = m$): $y_c = (c_1 + c_2 x)e^{m x}$
- Complex Conjugate Roots ($m = \alpha \pm i\beta$): $y_c = e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x)$
- Particular Integral ($y_p$): Accounts for the non-zero forcing function $f(x)$. It can be found using the method of undetermined coefficients (guessing a template solution based on the form of $f(x)$) or via the Variation of Parameters method for general functions.
Find the complete general solution of the second-order differential equation: $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$.
Show solution
The equation is a homogeneous second-order linear differential equation, so its solution consists entirely of the complementary function ($y = y_c$). Write down the corresponding auxiliary quadratic equation by replacing the derivatives with powers of $m$: \[ m^2 - 5m + 6 = 0 \] Factor the quadratic polynomial to find its roots: \[ (m - 2)(m - 3) = 0 \implies m_1 = 2, \quad m_2 = 3 \] The auxiliary equation has two real and distinct roots. Apply the corresponding solution template: \[ y(x) = c_1 e^{m_1 x} + c_2 e^{m_2 x} = c_1 e^{2x} + c_2 e^{3x} \] Final Answer: $y = c_1 e^{2x} + c_2 e^{3x}$.
Clairaut's Equation, Geometric Isoclines, and Orthogonal TrajectoriesTopic 2
- Clairaut's Equation: A first-order non-linear equation with the standard format $y = x p + f(p)$, where $p = \frac{dy}{dx}$. Its general solution is found by simply replacing the derivative term $p$ with an arbitrary constant $c$: $y = cx + f(c)$. This represents a family of straight lines. Differentiating with respect to $p$ yields the singular solution, which represents the envelope of these lines and contains no arbitrary constants.
- Orthogonal Trajectories: An orthogonal trajectory is a curve that intersects every member of a given family of curves at a right angle ($90^\circ$). To find the orthogonal family:
- Form the differential equation for the given family of curves: $F(x, y, \frac{dy}{dx}) = 0$.
- Replace the original derivative term $\frac{dy}{dx}$ with its negative reciprocal, $-\frac{dx}{dy}$, to enforce perpendicularity.
- Solve the resulting new differential equation to find the orthogonal curves.
- Slope Fields and Isoclines: Provide a method for visualizing solutions geometrically. An isocline for the equation $\frac{dy}{dx} = f(x,y)$ is a curve along which the slope of the solution path is constant ($f(x,y) = k$). Sketching these constant-slope line segments helps trace solution curves without solving the equation analytically.
Find the general solution of the following Clairaut equation: $y = x p + p^2 \quad \left(p = \frac{dy}{dx}\right)$.
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- Case 1: $\frac{dp}{dx} = 0 \implies p = c$ (an arbitrary constant). Substituting $p = c$ back into the original equation yields the general solution family of straight lines: \[ y = cx + c^2 \]
- Case 2: $x + 2p = 0 \implies p = -\frac{x}{2}$. Substituting this expression back into the original equation yields the singular solution envelope: \[ y = x\left(-\frac{x}{2}\right) + \left(-\frac{x}{2}\right)^2 = -\frac{x^2}{2} + \frac{x^2}{4} = -\frac{x^2}{4} \implies x^2 = -4y \]
Physical Modeling & Applications
Physical Applications and Modeling SystemsTopic 1
- Exponential Growth and Decay: Modeled by the first-order separable equation $\frac{dN}{dt} = kN$. If $k > 0$, the system experiences exponential growth (e.g., unconstrained population models); if $k < 0$, it experiences exponential decay (e.g., radioactive substance half-life tracking).
- Newton's Law of Cooling: States that the rate of change of the temperature $T$ of an object is directly proportional to the difference between its temperature and the surrounding ambient temperature $T_m$: \[ \frac{dT}{dt} = -k(T - T_m) \quad (\text{where } k > 0) \]
- Mixing Problems: Track the chemical salt content $S(t)$ in a fluid tank over time. The rate of change is modeled using mass balance mechanics: \[ \frac{dS}{dt} = \text{Rate in} - \text{Rate out} = (\text{Inflow Concentration} \times \text{Inflow Rate}) - \left(\frac{S(t)}{\text{Volume}} \times \text{Outflow Rate}\right) \]
- Projectile Motion with Air Resistance: Extends basic linear kinematics by adding a velocity-dependent drag force term ($F_{\text{drag}} = -cv$), transforming the system into a manageable first-order linear differential equation in terms of velocity.
A radioactive substance decays at a rate proportional to its mass. If the initial mass is $100\text{ mg}$ and it half-decays to $50\text{ mg}$ in exactly $20\text{ hours}$, find the expression for the mass remaining at any time $t$.
Show solution
Let $M(t)$ represent the mass of the radioactive substance at time $t$. The decay process is modeled by the differential equation: \[ \frac{dM}{dt} = -kM \quad (\text{where } k > 0) \] Separate the variables and integrate both sides: \[ \frac{1}{M} \, dM = -k \, dt \implies \ln M = -kt + C_1 \implies M(t) = e^{-kt + C_1} = C e^{-kt} \] Apply the initial condition ($M(0) = 100$) to find the constant $C$: \[ 100 = C e^0 \implies C = 100 \implies M(t) = 100 e^{-kt} \] Now apply the second condition ($M(20) = 50$) to solve for the decay constant $k$: \[ 50 = 100 e^{-20k} \implies e^{-20k} = \frac{1}{2} \implies -20k = \ln\left(\frac{1}{2}\right) = -\ln 2 \implies k = \frac{\ln 2}{20} \] Substitute the expression for $k$ back into the mass equation to find the final tracking formula: \[ M(t) = 100 e^{-\left(\frac{\ln 2}{20}\right)t} = 100 \left(e^{\ln 2}\right)^{-t/20} = 100 (2)^{-t/20} \] Final Answer: $M(t) = 100 \cdot 2^{-t/20}$.
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