Solve the given differential equation by separation of...

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used to model phenomena involving rates of change in fields like physics, biology, and engineering. Solving differential equations helps in determining the behavior of systems over time or space.

Separation of Variables

Separation of variables is a common method for solving first order differential equations. The technique involves algebraically manipulating the equation so that all terms involving one variable and its differential are on one side of the equation, and all terms involving the other variable and its differential are on the opposite side. This allows both sides to be integrated independently.

Algebraic Factorization

Algebraic factorization is used to simplify expressions by rewriting them as a product of simpler factors. In the context of differential equations, factorization can reveal structures in the equation that facilitate separation of variables or other solution methods by uncovering common factors that can be cancelled or grouped.

Integration of Rational Functions

Integration of rational functions involves integrating functions that can be expressed as the ratio of two polynomials. Once the differential equation has been separated, each side typically reduces to an integral of a rational function, which may require techniques such as partial fraction decomposition or substitution to resolve.

Constant of Integration

The constant of integration arises from the process of integrating indefinite integrals in the solution of differential equations. It represents the family of solutions that differ by a constant and is determined by initial or boundary conditions when available.

Transcript

00:01 And this question, we need to find a solution of this differential equation by separation of variables.

00:06 Let's first try to simplify this.

00:08 We have dy over dx and this is equal to xy plus 3x minus y minus 3 minus 3, minus y minus 3 over xy minus 2x plus 4y minus 8.

00:27 So this can be written as d .y over dx is equal to over here.

00:34 Let's try to factorize this.

00:36 So we take x out from here.

00:38 So we have y plus 3 and we take minus 1 out from here and we are left with y plus 3 again.

00:46 Over here we take x out.

00:48 So we are left with y minus 2.

00:50 And over here we take 4 out.

00:53 So we are again left with y minus 2.

00:56 So this can be rewritten as d .y over dx is equal to, if you take y plus three out, if we take y plus three as a common factor, we are left with x minus one.

01:08 And if we take y minus two as a common factor, we are left with x plus four over here.

01:17 So let's see.

01:18 Let's try to cross multiply this and bring the terms with the same variable over to one side.

01:24 So this can be written as y minus two over y plus three.

01:27 D .y and this is equal to x minus 1 over x plus 4 d x this is how it looks like...

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