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Calculus - Solving Differential Equations: Separation of Variables, Homogeneous, Linear form

Grade 12ICSE

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

🔑Concepts

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Differential Equation Fundamentals: A differential equation is an equation containing an unknown function and its derivatives. The order is the highest derivative present, while the degree is the power of that highest derivative (after removing radicals). Geometrically, a first-order differential equation defines a slope field where each point (x,y)(x, y) has a specific gradient assigned to it.

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Method of Separation of Variables: Used when the equation can be written as dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y). You visually rearrange the equation to group all yy terms with dydy on one side and all xx terms with dxdx on the other side, effectively 'splitting' the derivative symbol to allow for direct integration of both sides.

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Homogeneous Differential Equations: An equation of the form dydx=f(x,y)\frac{dy}{dx} = f(x, y) is homogeneous if f(x,y)f(x, y) is a homogeneous function of degree zero, meaning f(λx,λy)=f(x,y)f(\lambda x, \lambda y) = f(x, y). Visually, these equations describe slopes that remain constant along any straight line passing through the origin. They are solved using the substitution y=vxy = vx, which transforms them into a separable equation in terms of vv and xx.

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Linear Differential Equations (Leibniz Form): Equations of the form dydx+Py=Q\frac{dy}{dx} + Py = Q, where PP and QQ are functions of xx only. This form represents a system where the rate of change of yy depends linearly on yy itself. These are solved using an 'Integrating Factor' (IFIF) to make the left side a perfect derivative of a product.

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The Integrating Factor (IFIF): For the linear form, the IFIF is calculated as e∫Pdxe^{\int P dx}. Multiplying the entire equation by this factor allows the left side to be condensed into ddx(y⋅IF)\frac{d}{dx}(y \cdot IF). Visually, this factor 'scales' the equation to transform a non-exact differential into an exact one that is easily integrated.

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General vs. Particular Solutions: The general solution contains an arbitrary constant CC and represents a 'family of curves' filling the plane. A particular solution is obtained by using 'Initial Conditions' (e.g., y=y0y=y_0 when x=x0x=x_0) to find a specific value for CC. Geometrically, this corresponds to selecting one specific curve from the family that passes through a given point.

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Integration as the Inverse Process: Solving differential equations is essentially the search for the original function given its rate of change. On a graph, while the differential equation tells you the direction to move at any point, the solution gives you the path (the curve) that follows those directions consistently.

📐Formulae

Separable Form: ∫1h(y)dy=∫g(x)dx+C\int \frac{1}{h(y)} dy = \int g(x) dx + C

Homogeneous Substitution: y=vx⇒dydx=v+xdvdxy = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}

Linear Form (Standard): dydx+Py=Q\frac{dy}{dx} + Py = Q

Integrating Factor (IFIF): IF=e∫PdxIF = e^{\int P dx}

General Solution for Linear Form: y(IF)=∫(Q⋅IF)dx+Cy(IF) = \int (Q \cdot IF) dx + C

Linear Form (Alternative): dxdy+P1x=Q1⇒IF=e∫P1dy\frac{dx}{dy} + P_{1}x = Q_{1} \Rightarrow IF = e^{\int P_{1} dy}

General Solution (Alternative): x(IF)=∫(Q1⋅IF)dy+Cx(IF) = \int (Q_{1} \cdot IF) dy + C

💡Examples

Problem 1:

Solve the differential equation: dydx=1+y21+x2\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}

Solution:

Step 1: Separate the variables xx and yy. dy1+y2=dx1+x2\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} Step 2: Integrate both sides. ∫dy1+y2=∫dx1+x2\int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2} Step 3: Use the standard integral formula ∫11+z2dz=tan⁡−1z\int \frac{1}{1+z^2} dz = \tan^{-1} z. tan⁡−1y=tan⁡−1x+C\tan^{-1} y = \tan^{-1} x + C Step 4: (Optional) Express yy explicitly. y=tan⁡(tan⁡−1x+C)y = \tan(\tan^{-1} x + C) Using the identity tan⁡(A+B)\tan(A+B), this can be written as: y=x+tan⁡C1−xtan⁡Cy = \frac{x + \tan C}{1 - x \tan C} Let tan⁡C=K\tan C = K (another constant): y=x+K1−Kxy = \frac{x + K}{1 - Kx}

Explanation:

This is a classic 'Separation of Variables' problem. By moving all yy terms to the left and xx terms to the right, we transform the derivative equation into two simple independent integrals.

Problem 2:

Solve the linear differential equation: dydx+2y=e−x\frac{dy}{dx} + 2y = e^{-x}

Solution:

Step 1: Identify PP and QQ. Here, P=2P = 2 and Q=e−xQ = e^{-x}. Step 2: Calculate the Integrating Factor (IFIF). IF=e∫Pdx=e∫2dx=e2xIF = e^{\int P dx} = e^{\int 2 dx} = e^{2x} Step 3: Apply the general solution formula y(IF)=∫(Q⋅IF)dx+Cy(IF) = \int (Q \cdot IF) dx + C. y(e2x)=∫(e−x⋅e2x)dx+Cy(e^{2x}) = \int (e^{-x} \cdot e^{2x}) dx + C Step 4: Simplify the integrand using laws of exponents. y⋅e2x=∫exdx+Cy \cdot e^{2x} = \int e^{x} dx + C Step 5: Integrate and solve for yy. y⋅e2x=ex+Cy \cdot e^{2x} = e^{x} + C y=ex+Ce2x=e−x+Ce−2xy = \frac{e^x + C}{e^{2x}} = e^{-x} + Ce^{-2x}

Explanation:

This problem follows the 'Linear Form' procedure. The Integrating Factor e2xe^{2x} is used to collapse the left-hand side into a single derivative, making the equation solvable by direct integration of the right-hand side.