Find the general solution of each first-order linear differential equation. (d y)/(d x)-y cscx=s

Find the general solution of each first-order linear...

Key Concepts

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations. It involves finding a function ?(x) = exp(?P(x)dx) that, when multiplied by the entire equation, transforms the left-hand side into the derivative of a product of ?(x) and y. This simplifies the equation into a form that can be directly integrated to find the general solution.

First-Order Linear Differential Equations

These are differential equations of the form dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x. They are called first order because they involve only the first derivative of the unknown function y, and linear because the function and its derivative appear to the first power and are not multiplied together.

Standard Form and Coefficient Identification

To solve a first-order linear differential equation, it is essential to rewrite it in the standard form dy/dx + P(x)y = Q(x). This rearrangement allows for the clear identification of the function P(x), which multiplies y, and the non-homogeneous term Q(x), setting the stage for the integrating factor method.

Trigonometric Manipulation in Differential Equations

When the coefficients or non-homogeneous terms in a differential equation involve trigonometric functions, understanding trigonometric identities and manipulation becomes crucial. This allows the transformation and simplification of the equation, aiding in the identification of an integrating factor and the subsequent integration steps necessary for finding the general solution.

Transcript

00:01 Hello, the question is taken from differential equation and where we need to solve this differential equation or we need to evaluate the general solution of this first order linear differential equation.

00:13 So linear differential equation is dy over d x minus y cosecant x is equal to sine of 2x.

00:30 So this equation resembled with the normal differential equation of kind, dy over d x, p, of x into y is equal to q of x okay so how we can solve this kind of equation by evaluating the integrated factor and that integrated factor is integration e to the px d x if we compare these two equation p x is minus cosecant x so integrated factor eventually e to the power minus cosecant x d x and that will be equal to e to the power loch to the base e cosecant x plus quote x since the base is same so we get this is equal to cosecant x plus kote of x that is the required value of integrated factor so let us multiply this integrated factor throughout the equation so we get sorry cosecant x plus cortex into d y over d x minus cosecant x into cosecant x plus cortex into sine x x x plus cortex into y is equal to sine 2x into cosecant x plus cortex so left -hand side of the equation become d over d x y into cosecant x plus cortex and right -hand side become sine of 2x is 2 sine x co -x co -s -x and co -secent x is 1 by sine of x so sine x will be cancelled out we get 2 course of x plus yes and cortex is cos x over sine x so sine x will be cancelled out so we get two cos square x okay so taking d x to the right hand side we get d of y cose second x plus cortex integration and integrating and that is equal to two course of x plus x can be written as co square x 1 plus cos 2x divided by 2 into d x plus c c is the constant of integration so left -hand side become y into c second x plus quart of x that is equal to 2 into integration of course of x is equal to sine of x plus integration of 1 by 2 is 1 x by 2 and integration of course of 2 x is sine 2x divided by 4 plus c okay so taking this cosecant x plus cortex to the right hand side so cosec and x plus cortex is 1 plus cos x over sine x so y will be taking to inside we get sine x divided by sine x to sine x sorry sorry sorry taking lcm also to the 4 so that become 2 4x plus 4x because 2 will be cancelled out but but this 4 will come here 1 second 4x plus 2 x into x 2 is 4x and that will be sign 2x by 2 so that will be sign of 2x divided by 2 is in the lcm 1 second let me check this once again so first i will take the lcm and then do other thing so before that i can write it here as 1 plus o's of x divided by sine of x is equal to 2 into 4 sine x plus 2x plus sine x plus sine 2x plus c divided by 4...

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