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First, let's consider the following initial value of problem.

y double prime, plus 2y prime,

plus y is equal to e to the negative x over x squared plus one.

That's the governing differential equation.

And the given initial conditions are y(0) is equal to one half,

and y prime of zero is equal to negative one half.

First, note that in this problem,

the given g of x,

this input function, e to the negative x over x squared plus one.

It has no differential polynomial annihilator.

So even though the differential polynomial,

differential equation is set,

has a constant coefficients, but the input function has

no differential polynomial annihilator.

And that means we cannot apply the method of

undetermined coefficients to this problem to find its particular solution.

So let us help here could be another method,

the variation of parameters,

to find this particular solution.

First, let's try to find the complementary solution.

Think about corresponding homogeneous problem,

y double prime plus 2y prime,

plus y is equal to zero.

Its characteristic equation is r squared + 2r + 1,

or (r+1) squared is equal to zero.

So it has one double root,

r is equal to minus one.

And that means you have two linear independent solution of this homogeneous problem.

One is that e to the negative x,

and the other one y2 is equal to x times e to the negative x.

They are y1 and y2.

The Wronskian is y1 e to the minus x,

y2 x times e to the minus x.

There y one prime,

minus e to the minus x,

there y2 prime, derivative of this,

that is e to the negative x,

minus x times e to the negative x.

This determinant, you will get e to the negative 2x.

That is a capital W. Then by the variation of parameters,

we are looking for a particular solution of this problem

in the form of u1 times y1 plus u2 times y2,

where the u1 x and u2 x are unknown functions,

but they satisfy this simultaneous equation.

Say, y1 u1 prime + y2 u2 prime is equal to zero.

And y1 prime u1 prime plus y2 u2 prime is equal to g of x.

That is equal to e to the negative x over x squared plus one.

Solving this simultaneous equation,

you will get u1 prime is equal

to negative y2 g. Can you remind of the general formula?

Let me write it here.

u1 prime is equal to negative g of y2 over Wronskian.

On the other hand, u2 prime is equal to g of y1 over Wronskian.

So applied in this case,

what is the y2?

y2 is equal to x times e to the negative x.

What is g? g is equal to e to the negative x over x squared plus one.

What is W?

W is equal to e to the negative 2x.

So compute this quantity, okay?

Then, you will get u1 prime is equal to negative x over x squared plus one.

On the other hand,

u2 prime, g is equal to this one.

And y_sub_1 is equal to u to the negative x,

and the capital W is equal to e to the negative 2x.

So that u2 prime will be equal to,

in this case, one over x squared plus one.

Integrating, what is u1?

u1 is the antiderivative of minus x over x squared plus one,

which is equal to negative one half times log x squared plus one.

You may add the possibly integrating constant,

which we are not taking to be zero

without losing any generality.

For the u2 case,

u2 must to be antiderivative of one over x squared plus one,

which is the arctangent x plus some another arbitrary integrating constant,

which we may take to be zero.

So our choice of u1 and u2 are u1 is equal to negative one half log x squared plus one,

and u2 is equal to arc tangent of x.

Then the particular solution we are looking for is,

so u1 times, what is yp again?

yp, this is equal to u1 y1 + u2 y2.

So we already have y1 and y2.

And we just obtained u1 and u2.

So computing this quantity,

you will get particular solution will be minus one over two,

e to the negative x, times the log x squared plus one,

plus x e to the negative x times arc tangent x.

This is the particular solution of given non-homogenous equation right here.

That's the particular solution.

So its general solution will be y is equal to

complementary solution plus particular solution.

Now applied finally the two initial condition,

say y of zero is equal to one half,

and y prime zero is equal to negative one half.

y of zero computing from this one,

you will get c1.

y prime of zero computing from this expression,

you will get that is equal to negative c1 plus c2,

must be equal to negative one half.

Solving this simultaneous equation for c1 and c2,

you will get c1 is equal to one half,

and c2 is equal to zero.

So finally, you will get the solution to the,

because the initial value problem will be,

one half e to the negative x minus one half e to

the negative x times log x squared plus one,

plus x e to the negative x and arc tangent x.