# Homogeneous Linear D.E. Solutions: Step-by-Step Guide | Urgent Help

• rygza
In summary, the given linear equation is said to be homogeneous since Q(x)=0. The trivial solution is y=0, and if y=y1(x) is a solution and k is a constant, then y=ky1x is also a solution. If y=y1x and y=y2x are solutions, then y=y1x + y2x is also a solution. To solve the problems, you can use the fact that constants can be pulled out and applied after differentiation, and that combining solutions will result in a solution as long as each individual term is a solution.
rygza

## Homework Statement

A linear equation in form:

dy/dx + P(x)y = 0 is said to be homogeneous since Q(x)=0.

a) show that y=0 is a trivial solution (wasn't even taught what a trivial solution is)
b) show that y=y1(x) is a solution and k is a constant, then y=ky1x is also a solution.
c) show that if y=y1x and y=y2x are solutions, then y=y1x + y2x is a solution

I don't even know how to start this problem. For part a) i simply plugged in 0 for y and got dy/dx=0 . doesn't seem right. then i tried separation of variable and got stuck at

(1/y)dy=-P(x)dx

can someone please guide me through? i have about three other problems like this and i haven't got a clue how to solve them.

p.s. that's y(sub1) and y(sub2)

you're not solving anything specific, since you don't really know about your functions. you're just proving general cases.

the trivial solution is just y=0. that's certainly true, dy/dx of 0 = 0, and P(x)y=P(x)0=0.

so, if you know that y1 is a solution, what do we know about constants? we can pull them out and apply them after we differentiate. so that's just like multiplying each term by k, and k0 = 0.

so we know that y1 and y2 are solutions in themselves. if you plug in y1 + y2 as a solution, well,
dy/dx (y1 + y2) = dy/dx y1 + dy/dx y2.
and P(x)(y1 + y2) = P(x)y1 + P(x) y2.
so dy/dx (y1 + y2) + P(x)(y1 + y2) = dy/dx y1 + dy/dx y2 + P(x)y1 + P(x) y2 = dy/dx y1 + P(x)y1 + dy/dx y2 + P(x)y2.
we know that dy/dx y1 + P(x)y1 = 0, and we know that dy/dx y2 + P(x)y2 = 0, since they're both solutions, so then dy/dx y1 + P(x)y1 + dy/dx y2 + P(x)y2 = 0.
i hope that's right, I'm a bit rusty with this stuff.

## 1. What is a homogeneous linear differential equation?

A homogeneous linear differential equation is a type of mathematical equation that involves an unknown function and its derivatives. It is considered homogeneous because all terms of the equation have the same degree with respect to the unknown function.

## 2. How do I solve a homogeneous linear differential equation?

To solve a homogeneous linear differential equation, you can use the method of separation of variables or the method of undetermined coefficients. Alternatively, you can also use the characteristic equation method or the method of variation of parameters.

## 3. What is the difference between a homogeneous and non-homogeneous linear differential equation?

In a homogeneous linear differential equation, all terms have the same degree with respect to the unknown function, whereas in a non-homogeneous linear differential equation, the terms have different degrees. This means that a non-homogeneous equation will have an additional term that depends on the independent variable.

## 4. Are there any real-world applications of homogeneous linear differential equations?

Yes, homogeneous linear differential equations have many applications in physics, engineering, and economics. For example, they can be used to model the growth of a population, the movement of a pendulum, or the spread of a disease.

## 5. What is the role of initial conditions in solving a homogeneous linear differential equation?

Initial conditions are used to determine the specific solution to a homogeneous linear differential equation. They are usually given in the form of values for the unknown function and its derivatives at a specific point. These initial conditions are essential in solving the differential equation and finding the unique solution.

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