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Differential Equations - Solution of linear differential equation of the type dy/dx + py = q and dx/dy + px = q

Grade 12CBSE

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

🔑Concepts

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A first-order linear differential equation is an equation where the dependent variable and its derivative appear only in the first power and are not multiplied together. Visually, this ensures that the rate of change is a linear function of the variable itself, leading to smooth, continuous growth or decay curves when plotted.

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The first standard form is dydx+Py=Q\frac{dy}{dx} + Py = Q, where PP and QQ are either constants or functions of xx only. In this form, yy is the dependent variable and xx is the independent variable. On a graph, this describes how the vertical position yy changes relative to the horizontal position xx.

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To solve the first form, we use an Integrating Factor (I.F.), defined as I.F.=e∫PdxI.F. = e^{\int P dx}. Multiplying the entire equation by this factor transforms the left-hand side into the exact derivative of the product (y⋅I.F.)(y \cdot I.F.). Conceptually, the I.F. acts as a scaling factor that 'collapses' the sum of two terms into a single differentiable unit.

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The second standard form is dxdy+Px=Q\frac{dx}{dy} + Px = Q, where PP and QQ are constants or functions of yy only. Here, the roles are reversed: xx is the dependent variable and yy is the independent variable. This is often used when the equation is easier to solve by viewing the rate of change along the y-axis.

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The Integrating Factor for the second form is I.F.=e∫PdyI.F. = e^{\int P dy}. The solution follows the structure x⋅I.F.=∫(Q⋅I.F.)dy+Cx \cdot I.F. = \int (Q \cdot I.F.) dy + C. This mirrors the first form but integrates with respect to yy, representing a horizontal shift or mapping relative to the vertical axis.

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The General Solution of a linear differential equation represents a 'family of curves.' Each different value of the arbitrary constant CC produces a distinct curve on the Cartesian plane. If an initial condition (x0,y0)(x_0, y_0) is provided, we identify one specific member of this family that passes through that point, known as the Particular Solution.

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Logarithmic Identity: A crucial simplification used in these problems is eln⁡f(x)=f(x)e^{\ln f(x)} = f(x). This is frequently used after integrating PP to simplify the I.F. from an exponential form into a polynomial or trigonometric function, which makes the subsequent integration of Q⋅I.F.Q \cdot I.F. manageable.

📐Formulae

Type 1: dydx+Py=Q\frac{dy}{dx} + Py = Q (where P,QP, Q are functions of xx)

Integrating Factor (Type 1): I.F.=e∫PdxI.F. = e^{\int P dx}

General Solution (Type 1): y⋅(I.F.)=∫(Q⋅I.F.)dx+Cy \cdot (I.F.) = \int (Q \cdot I.F.) dx + C

Type 2: dxdy+Px=Q\frac{dx}{dy} + Px = Q (where P,QP, Q are functions of yy)

Integrating Factor (Type 2): I.F.=e∫PdyI.F. = e^{\int P dy}

General Solution (Type 2): x⋅(I.F.)=∫(Q⋅I.F.)dy+Cx \cdot (I.F.) = \int (Q \cdot I.F.) dy + C

Property of Exponents: eln⁡(g(x))=g(x)e^{\ln(g(x))} = g(x)

💡Examples

Problem 1:

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

Solution:

  1. Identify PP and QQ: Here, P=2P = 2 and Q=e−xQ = e^{-x}. \n2. Calculate the Integrating Factor: I.F.=e∫2dx=e2xI.F. = e^{\int 2 dx} = e^{2x}. \n3. Apply the general solution formula: y⋅e2x=∫(e−x⋅e2x)dx+Cy \cdot e^{2x} = \int (e^{-x} \cdot e^{2x}) dx + C. \n4. Simplify the integrand: y⋅e2x=∫exdx+Cy \cdot e^{2x} = \int e^x dx + C. \n5. Integrate: y⋅e2x=ex+Cy \cdot e^{2x} = e^x + C. \n6. Solve for yy: y=e−x+Ce−2xy = e^{-x} + Ce^{-2x}.

Explanation:

This is a Type 1 linear differential equation. We find the I.F. to combine the yy and dy/dxdy/dx terms into a single derivative of a product, then integrate the right side.

Problem 2:

Solve the differential equation: xdydx−y=2x2x \frac{dy}{dx} - y = 2x^2

Solution:

  1. Divide by xx to get standard form: dydx−1xy=2x\frac{dy}{dx} - \frac{1}{x}y = 2x. \n2. Identify P=−1xP = -\frac{1}{x} and Q=2xQ = 2x. \n3. Find I.F.: I.F.=e∫−1xdx=e−ln⁡x=eln⁡x−1=1xI.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = e^{\ln x^{-1}} = \frac{1}{x}. \n4. Apply formula: y⋅1x=∫(2x⋅1x)dx+Cy \cdot \frac{1}{x} = \int (2x \cdot \frac{1}{x}) dx + C. \n5. Simplify: yx=∫2dx+C\frac{y}{x} = \int 2 dx + C. \n6. Integrate: yx=2x+C\frac{y}{x} = 2x + C. \n7. Final solution: y=2x2+Cxy = 2x^2 + Cx.

Explanation:

Before identifying PP and QQ, the coefficient of dy/dxdy/dx must be 1. After normalizing the equation, we use the logarithmic identity to simplify the Integrating Factor.