Question Video: Solving First-Order First-Degree Linear Differential Equations | Nagwa Question Video: Solving First-Order First-Degree Linear Differential Equations | Nagwa

# Question Video: Solving First-Order First-Degree Linear Differential Equations Mathematics • Higher Education

Is the differential equation (dπ¦/dπ₯) + π₯β(π¦) = π₯Β² linear?

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### Video Transcript

Is the differential equation dπ¦ by dπ₯ plus π₯ root π¦ equals π₯ squared linear?

A linear differential equation is one which can be expressed as a linear polynomial of the unknown function, in this case π¦ and its derivatives. What this means is that the only powers of the unknown function and each derivative that appear in the equation are one or zero if the equation does not contain that order derivative. And also each derivative and the function itself are multiplied by functions of π₯ only.

So, for example, the equation two times dπ¦ by dπ₯ plus four π₯π¦ equals three π₯ would be an example of a linear differential equation. Because the power of both π¦ and dπ¦ by dπ₯ is one, and theyβre each multiplied by a function of π₯ only. Whereas the equation two times dπ¦ by dπ₯ plus four π₯ over π¦ equals three π₯ is nonlinear as in the second term, the power of π¦ is negative one. The equation four π₯ d two π¦ by dπ₯ squared plus two π¦ dπ¦ by dπ₯ equals seven is also nonlinear as in the second term, we see that dπ¦ by dπ₯ is multiplied by a function of π¦, not a pure function of π₯.

More formally, we can say that a differential equation is linear if it can be expressed in the form shown on the screen. Each πth order derivative of π¦ and the function π¦ itself is multiplied by a polynomial in π₯ only. So, letβs consider the differential equation that weβve been given. And we can see that it includes the square root of π¦. Now, another way of expressing the square root of π¦ is as π¦ to the power of one-half. And hence, this differential equation is nonlinear, as the power of π¦ is not equal to one.

Now, notice that it isnβt the presence of the π₯ squared term on the right-hand side that makes this differential equation nonlinear. π₯ is the independent variable in this equation. And it is only the powers of the dependent variable and its derivatives, thatβs π¦ and dπ¦ by dπ₯ and so on, that must all be equal to one in order to make the equation linear.

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