Question

find the particural integral (D²-4)y=e^-4×​

Answer 1

Answer:

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Step-by-step explanation:

What is the particular integral of (D^2-4D+4) y=e^x?

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What is the particular integral of (D^2-4D+4) y=e^x?

Before rushing in to work out the particular integral, we must first take time to find the solution of the complementary function (D2−4D+4)y=0 .

First, we need to find the roots of the auxiliary equation D2−4D+4=0⇒D=2 (a repeated root).

Thus the solution of the Complementary Function is y=(Ax+B)e2x

where A and B can have any value.

Now for the particular integral. As we want to end up with ex , this suggests using:

y=cex⇒y′=cex⇒y"=cex

Putting these into our original equation, we have: cex−4cex+4cex=ex

Simplifying the left side: cex=ex⇒c=1

Thus the Particular Integral is y=ex

Adding the Particular Integral to the solution of the Complementary Function gives us the General Solution: y=(Ax+B)e2x+ex

You might be wondering why we just didn’t go straight to evaluating the Particular Integral. Well, consider the slightly different equation (D2−4D+3)y=ex

Let’s just jump in and assume that the Particular Integral is in the form:

y=cex⇒y′=cex⇒y"=cex

The equation thus becomes: cex−4cex+3cex=ex

Simplifying: 0=ex

What gives! Well, let’s look at the Complementary Function: (D2−4D+3)y=0

The auxiliary equation is: D2−4D+3=0⇒D=1 or D=3

The solution to the Complementary Function is thus: y=Aex+Be3x

As y=Aex has already been taken by the solution for the Complementary Function, you can’t use it for the Particular Integral. Instead, we have to ‘bump it up’ by multiplying by x , giving us:

y=cxex⇒y′=(cx+c)ex⇒y"=(cx+2c)ex

Putting this into the equation: (cx+2c)ex−4(cx+c)ex+3cxex=ex

Simplifying the left side: −2cex=ex⇒c=−12

Thus the Particular Integral would be: y=−12xex