When do you use integrating factor?

The usage of integrating factor is to find a solution to differential equation. Integrating factor is used when we have the following first order linear differential equation. It can be homogeneous(when Q(x)=0) or non homogeneous. where P(x) & Q(x) is a function of x.

Which is the linear differential equation?

The linear differential equation is of the form dy/dx + Py = Q, where P and Q are numeric constants or functions in x. It consists of a y and a derivative of y. The differential is a first-order differentiation and is called the first-order linear differential equation. This linear differential equation is in y.

How do you find the specific solution of a first order differential equation?

Steps

1. Substitute y = uv, and.
2. Factor the parts involving v.
3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
4. Solve using separation of variables to find u.
5. Substitute u back into the equation we got at step 2.
6. Solve that to find v.

What is meant by particular solution?

Definition of particular solution : the solution of a differential equation obtained by assigning particular values to the arbitrary constants in the general solution.

How do you find the general solution of a linear differential equation?

follow these steps to determine the general solution y(t) using an integrating factor:

1. Calculate the integrating factor I(t). I ( t ) .
2. Multiply the standard form equation by I(t). I ( t ) .
3. Simplify the left-hand side to. ddt[I(t)y]. d d t [ I ( t ) y ] .
4. Integrate both sides of the equation.
5. Solve for y(t). y ( t ) .

What is a particular equation?

: the solution of a differential equation obtained by assigning particular values to the arbitrary constants in the general solution.

What are the particular and total solution?

The total solution or the general solution of a non-homogeneous linear difference equation with constant coefficients is the sum of the homogeneous solution and a particular solution. If no initial conditions are given, obtain n linear equations in n unknowns and solve them, if possible to get total solutions.

What is the actual solution of C1 and C2?

Now, apply the initial conditions to these. Solving this system gives c 1 = 2 c 1 = 2 and c 2 = 1 c 2 = 1. The actual solution is then. This will be the only IVP in this section so don’t forget how these are done for nonhomogeneous differential equations!

What is a particular solution to the differential equation?

A particular solution to the differential equation is then, Notice that if we had had a cosine instead of a sine in the last example then our guess would have been the same. In fact, if both a sine and a cosine had shown up we will see that the same guess will also work.

Why do we say “a particular solution” instead of the particular solution?

Notice in the last example that we kept saying “a” particular solution, not “the” particular solution. This is because there are other possibilities out there for the particular solution we’ve just managed to find one of them. Any of them will work when it comes to writing down the general solution to the differential equation. Speaking of which…

What is the differential equation m dv dt = mg?

Thus the differential equation m dv dt = mg is amathematical modelcorresponding to a falling object. To solve the differential equation, cancel the mass and note that v is an antiderivative of the constant g; thus v = gt + C, where C is an arbitrary constant.