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Calculus - Differential Equations: Order, Degree, General and Particular Solutions

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Differential Equation (DE) is an equation involving an independent variable xx, a dependent variable yy, and its derivatives like dydx\frac{dy}{dx}, d2ydx2\frac{d^2y}{dx^2}, etc. Visually, a DE defines a relationship between the coordinates of a point on a curve and the slope or curvature of the curve at that point.

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The Order of a differential equation is the order of the highest-order derivative appearing in the equation. For example, in d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0, the order is 22. Visually, higher orders correspond to higher-level geometric properties like concavity or rate of change of curvature.

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The Degree of a differential equation is the power or exponent of the highest-order derivative, provided the equation is expressed as a polynomial in derivatives (i.e., derivatives are free from radicals and fractional powers). If the highest derivative is inside a function like sin⁡(dydx)\sin(\frac{dy}{dx}) or edydxe^{\frac{dy}{dx}}, the degree is not defined.

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A General Solution of a differential equation of order nn is a solution that contains nn independent arbitrary constants (e.g., C1,C2C_1, C_2). Visually, the general solution represents a 'family of curves' covering a region of the Cartesian plane, where each value of the constants generates a different curve in the family.

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A Particular Solution is a solution obtained by assigning specific values to the arbitrary constants in the general solution, usually based on given initial or boundary conditions like y=y0y = y_0 when x=x0x = x_0. Visually, this corresponds to selecting a single, unique curve from the entire family that passes through a specific point (x0,y0)(x_0, y_0).

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To form a differential equation from a given family of curves (general solution), we differentiate the equation as many times as there are arbitrary constants and then eliminate those constants using the original and derived equations. The resulting DE represents the geometric property shared by every member of that family.

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A differential equation is Linear if the dependent variable yy and its derivatives occur only in the first degree and are not multiplied together. Otherwise, it is Non-linear. Graphically, linear differential equations often describe systems where the response is proportional to the input.

📐Formulae

General form of an nthn^{th} order DE: F(x,y,dydx,d2ydx2,...,dnydxn)=0F(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, ..., \frac{d^ny}{dx^n}) = 0

Condition for Degree: The equation must be a polynomial in derivatives, e.g., a0(dnydxn)k+a1(dnydxn)k−1+...=0a_0(\frac{d^ny}{dx^n})^k + a_1(\frac{d^ny}{dx^n})^{k-1} + ... = 0, where kk is the degree.

Relationship between Order and Constants: Number of arbitrary constants in General Solution=Order of the DE\text{Number of arbitrary constants in General Solution} = \text{Order of the DE}

Particular Solution condition: y=f(x,c1,c2)y = f(x, c_1, c_2) becomes particular when c1,c2c_1, c_2 are fixed values.

💡Examples

Problem 1:

Determine the order and degree (if defined) of the differential equation: (d2ydx2)3+(dydx)2+sin⁡(dydx)+1=0\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + \sin\left(\frac{dy}{dx}\right) + 1 = 0.

Solution:

  1. Identify the highest order derivative: The highest derivative is d2ydx2\frac{d^2y}{dx^2}, so the Order = 22. \n2. Check for polynomial form in derivatives: The equation contains a term sin⁡(dydx)\sin\left(\frac{dy}{dx}\right). Since it is not a polynomial in terms of its derivatives, the Degree is not defined.

Explanation:

Order is always defined by the highest derivative present. However, because the derivative is an argument of a trigonometric function, the equation cannot be written as a standard polynomial in terms of derivatives, making the degree undefined.

Problem 2:

Verify that y=Aex+Be−xy = Ae^x + Be^{-x} is a general solution of the differential equation d2ydx2−y=0\frac{d^2y}{dx^2} - y = 0.

Solution:

  1. First derivative: dydx=ddx(Aex+Be−x)=Aex−Be−x\frac{dy}{dx} = \frac{d}{dx}(Ae^x + Be^{-x}) = Ae^x - Be^{-x}. \n2. Second derivative: d2ydx2=ddx(Aex−Be−x)=Aex+Be−x\frac{d^2y}{dx^2} = \frac{d}{dx}(Ae^x - Be^{-x}) = Ae^x + Be^{-x}. \n3. Substitute into the DE: LHS = d2ydx2−y=(Aex+Be−x)−(Aex+Be−x)=0\frac{d^2y}{dx^2} - y = (Ae^x + Be^{-x}) - (Ae^x + Be^{-x}) = 0. \n4. Since LHS = RHS, the given function is a solution.

Explanation:

To verify a solution, we differentiate the given function up to the order of the DE and substitute the expressions back into the equation. Since there are two arbitrary constants (AA and BB) and the DE is of order 2, this is the general solution.