how do I solve a differential equation explicitly?

Question

Say for example I had y' = 15-3y
How would I solve that explicitly?

Answer 1

That's a first-order, linear, nonhomogeneous equation.

First you'd solve the homogeneous equation:

y' + 3y = 0

The answer to that is Ae^(-3x), where A would be determined by boundary conditions (which you don't specify).

Then you need a particular solution. Try a constant, y=B. Then y' = 0.

0 = 15 - 3B, so B=5

So the answer would be y = Ae^(-3x) + 5

Where A is some constant.

Answer 2

frequently, an function is written explicitly while one variable is exactly in terms of yet another. One often looks via itself on the left edge of the equation, and the final component is composed of any algebraic expression alongside with (or possibly not alongside with) the different variable. The based variable (the single on the left) won't be able to seem everywhere on the final. y = x^2 + 2 . in this occasion, y is expressed as an exhibit function of x. x^2 - y = -2. Written this type, this is not an exhibit function. y remains seen to be an "implicit" function of x. In differential equations, you often have 2 variables interior the challenge, yet there can in undemanding terms be one self sufficient variable. the different letter represents a _function_ of the self sufficient variable. (Exception: in _partial_ differential equations, there might desire to be extra effective than one self sufficient variable, yet it is in general addressed in an fullyyt separate course.) to unravel "explicitly" ability to locate the _function_ that solves the equation. it is, resolve for the based variable (the single whose by-product is given interior the challenge). as an occasion, interior the equation: y' + t^2 * y = 3t^3, " y' " ability "the by-fabricated from y with admire to t." Then y is seen to be a function of t. to unravel explicitly, the respond might look something like this: y = ~~t~~~. it is, y is on one edge of the equation via itself. the different edge of the equation is any algebraic expression, possibly alongside with t. There could be no different y's interior the challenge. In smart notation, the respond might seem as though this: y(t) = ~~t~~~. quite than y', you have gotten " dy/dt" quite. if so, this is extra handy to verify it is the self sufficient variable (the t interior the denominator) and it is the function (y, a function of t).

Answer 3

dy/dx = 15 - 3y
dy/(15 - 3y) = dx
let u = 15 - 3y, du = -3dy
-(1/3)∫du/u = ∫dx
ln(u) = -3x + C
15 - 3y = e^(-3x + C)
y = 5 - (C/3)e^-3x

Answer 4

y' = 15-3y
dy/dx = 15-3y
dy/(15-3y) = dx

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