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Solve the given differential equation.$$\frac{d y}{d x}=\frac{y^{2}+x y+x^{2}}{x^{2}}$$

$y=x \tan (\ln |x|+c)$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 8

Change of Variables

Differential Equations

Missouri State University

Campbell University

Harvey Mudd College

University of Nottingham

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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Solve the given differenti…

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Solve the differential equ…

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so, miss problem. We've been given the differential equation on the left hand side and we've been asked to solve it. Now, at first glance, it looks like a first order homogeneous equation. So we will check this in the blue box by making the substitution X equals t X and y equals t y in the function on the right hand side, making this substitution, we get that the function becomes t squared. Why squad plus t squared x y plus t squared x squared and then this is all over. He's got that squared now for is a common factor of tea squad in every time so we can cancel this out between numerator and it's nominator. And this is simply for right hand side we began with. So the right hand side function is a function that is homogeneous of degree zero. Service is a homogeneous differential equation. Best means that we can make the substitution of B equals y over X. But more usefully, this is the same as y equals V X, because we most by breath, sides by X now for a function, we want to find d y the X, and this is solved using the product rule on this function y equals V X. So doing the product rule, we find d y. The X is equal to V plus x times DV the X. So this is an important substitution to keep in mind. So we'll put that on the left hand side and then the right hand side. Or simply substitute y equals V X into the right hand side function here. So a differential equation now becomes fee plus X DVD X equal to V squared X squared, plus the X squared plus X squared all over X squared. Now again, this is a common factor of ex lead in every time in the numerator and denominator service cancels out and we are left with the sweat plus V plus One. What we can do some further simplification by noticing On the left hand side we have a common factor of the with the on the right hand side. And if we subtract this from both sides, we get a separable differential equation of X DVD X is equal to B squared plus one. This is several equation we're going to want to solve. We do this by separating the variables a veto, the left hand side and the ex variables to the right hand side, which gives us that one over B squared plus one. TV is equal to one over x the x onda. We will get a solution for this by integrating both sides. Integrating. This uses known integration solutions because one over V squared plus one is integrated to the art town. That the on the integral of one of the X is the natural log of X, and then we'll add the integration constancy to the right hand side. To get the solution for V, we therefore need to take the town of both sides, which gives us v equals b tan, the natural log of X plus C and this is all solution for V. But remember, we've been asked for the solution for why and thanks to a variable substitution of why equals X times V, we get that wine is simply x times the town the natural Oh, the vax plus C

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