krit.club logo

Differential Equations - Solution by method of separation of variables

Grade 12CBSE

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

🔑Concepts

Definition of Variable Separable Form: A first-order differential equation is said to be in variable separable form if it can be expressed in the form dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y). Visually, this means the rate of change can be factored into a component that depends only on the horizontal position xx and a component that depends only on the vertical position yy.

The Separation Process: The fundamental step involves algebraically rearranging the equation so that all terms involving yy (including dydy) are on one side of the equation and all terms involving xx (including dxdx) are on the other side, resulting in 1g(y)dy=f(x)dx\frac{1}{g(y)} dy = f(x) dx. This conceptually decouples the variables, allowing for independent integration.

Integration and the Arbitrary Constant: Once separated, we apply the integral sign to both sides: 1g(y)dy=f(x)dx+C\int \frac{1}{g(y)} dy = \int f(x) dx + C. The constant of integration CC is essential because it accounts for the fact that many functions can have the same derivative. Visually, the result describes a whole family of curves rather than a single path.

General Solution vs. Family of Curves: The resulting equation G(y)=F(x)+CG(y) = F(x) + C is called the General Solution. Geometrically, this represents an infinite set of curves in the xyxy-plane that do not intersect. Changing the value of CC effectively 'shifts' the curve across the plane without changing its fundamental slope characteristics at any given (x,y)(x, y).

Particular Solution and Initial Conditions: A Particular Solution is obtained when a specific point (x0,y0)(x_0, y_0) is provided, known as an initial condition. By substituting these coordinates into the general solution, we calculate a specific value for CC. Graphically, this 'picks out' the single unique curve from the entire family that passes through that specific point.

Handling Logarithmic Results: In many separation problems involving 1ydy\frac{1}{y} dy, the result is lny=f(x)+C\ln|y| = f(x) + C. To simplify, we often express the constant as lnk\ln|k| so that the entire equation can be simplified using logarithmic properties, ultimately leading to an explicit form y=kef(x)y = k \cdot e^{f(x)}.

Substitution for Non-Separable Forms: Some equations like dydx=f(ax+by+c)\frac{dy}{dx} = f(ax + by + c) are not initially separable. However, by using a visual transformation where we let v=ax+by+cv = ax + by + c, we can differentiate vv with respect to xx and substitute back into the original equation to create a new differential equation in vv and xx that is separable.

📐Formulae

dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y)

dyg(y)=f(x)dx\frac{dy}{g(y)} = f(x)dx

1g(y)dy=f(x)dx+C\int \frac{1}{g(y)} dy = \int f(x) dx + C

dydx=h(x)k(y)    k(y)dy=h(x)dx+C\frac{dy}{dx} = \frac{h(x)}{k(y)} \implies \int k(y) dy = \int h(x) dx + C

Initial Condition: y(x0)=y0y(x_0) = y_0

💡Examples

Problem 1:

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

Solution:

Step 1: Separate the variables by moving all yy terms to the left and xx terms to the right: 11+y2dy=11+x2dx\frac{1}{1 + y^2} dy = \frac{1}{1 + x^2} dx. Step 2: Integrate both sides: 11+y2dy=11+x2dx\int \frac{1}{1 + y^2} dy = \int \frac{1}{1 + x^2} dx. Step 3: Apply standard integration formulas: tan1(y)=tan1(x)+C\tan^{-1}(y) = \tan^{-1}(x) + C. Step 4: (Optional) Simplify the expression using the identity tan1Atan1B=tan1(AB1+AB)\tan^{-1} A - \tan^{-1} B = \tan^{-1}(\frac{A-B}{1+AB}). tan1(y)tan1(x)=C    tan1(yx1+xy)=C\tan^{-1}(y) - \tan^{-1}(x) = C \implies \tan^{-1}(\frac{y-x}{1+xy}) = C. Step 5: Let tan(C)=A\tan(C) = A (another constant), then yx1+xy=A\frac{y-x}{1+xy} = A or yx=A(1+xy)y-x = A(1+xy).

Explanation:

This is a classic variable separable problem. We isolate the inverse trigonometric forms and integrate. The constant CC represents a family of curves that look like rotated hyperbolas or similar transcendental curves.

Problem 2:

Find the particular solution of dydx=4xy2\frac{dy}{dx} = -4xy^2 given that y=1y = 1 when x=0x = 0.

Solution:

Step 1: Separate the variables: 1y2dy=4xdx\frac{1}{y^2} dy = -4x dx. Step 2: Rewrite the left side for easier integration: y2dy=4xdxy^{-2} dy = -4x dx. Step 3: Integrate both sides: y2dy=4xdx    y11=4x22+C\int y^{-2} dy = \int -4x dx \implies \frac{y^{-1}}{-1} = \frac{-4x^2}{2} + C. Step 4: Simplify: 1y=2x2+C-\frac{1}{y} = -2x^2 + C. Step 5: Use the initial condition x=0,y=1x=0, y=1 to find CC: 11=2(0)2+C    C=1-\frac{1}{1} = -2(0)^2 + C \implies C = -1. Step 6: Substitute CC back into the general equation: 1y=2x21-\frac{1}{y} = -2x^2 - 1. Step 7: Multiply by 1-1 and solve for yy: 1y=2x2+1    y=12x2+1\frac{1}{y} = 2x^2 + 1 \implies y = \frac{1}{2x^2 + 1}.

Explanation:

First, we find the general solution by separating yy and xx and integrating power functions. Then, we use the specific coordinate (0,1)(0, 1) to find the exact value of the constant CC, resulting in a unique bell-shaped curve.