Chapter 4: Q43E (page 151)

Consider the differential equation $a y'' + b y' + c y = e^{k x}$ , where a, b, c, and k are constants. The auxiliary equation of the associated homogeneous equation is $a m^{2} + b m + c = 0$ .
(a) If k is not a root of the auxiliary equation, show that we Can find a particular solution of the form $y_{g} = A e^{k x}$ , where $A = 1 / (a k^{2} + b k + c)$
(b) If k is a root of the auxiliary equation of multiplicity one, show that we can find a particular solution of the form $y_{p} = A x e^{k x}, w h e r e A = 1 / (2 a k + b)$ . Explain how we know that $k \neq - b / (2 a)$
(c) If k is a root of the auxiliary equation of multiplicity two, show that we can find a particular solution of the form $y = A x^{2} e^{k x}, w h e r e A = 1 / (2 a)$

Short Answer

Expert verified

a) Proved

b) Proved

c) Proved

Step by step solution

Given information

We have the second order differential equation

$a y'' + b y' + c y = e^{k x} \cdot \cdot \cdot \cdot \cdot \cdot (1)$

And its auxiliary equation is

$a m^{2} + b m + c = 0$

$y_{g} = A e^{k x}, A = 1 / (a k^{2} + b k + c)$

$y_{p} = A x e^{k x}, w h e r e A = 1 / (2 a k + b)$

$y = A x^{2} e^{k x}, w h e r e A = 1 / (2 a)$

Step 2(a): Show that a particular solution of the form yg=Aekx , where A=1/(ak2+bk+c)

If is not a root of its auxiliary equation, then we can have the homogeneous solution of our given differential equation as

$y_{h} = c_{1} e^{m_{1} x} + c_{2} e^{m_{2} x} a n d m_{1}, m_{2} \neq k$

Now we have to obtain the particular solution by assuming that $y_{p} = A e^{k x}$ , where the function $g (x) = A e^{k x}$ , and we can obtain the value of A as the following technique:

Differentiate the assumption with respect to x, and then we have

$y_{p}' = k A e^{k x} . . . . . . (2) y_{p}'' = k^{2} A e^{k x} . . . . . (3)$

Now we have to substitute with and equations (2) and (3) into equation (1), then we can have

$a k^{2} A e^{k x} + b K A e^{k x} + c A e^{k x} = e^{k x} a k^{2} A + b K A + c A = 1 A (a k^{2} + b k + c) = 1$

Then we obtain

$A = \frac{1}{a k^{2} + b k + c}$

Hence, it is shown that we Can find a particular solution of the form role="math" localid="1663924426535" $y_{g} = A e^{k x}$ , where $A = 1 / ({ak}^{2} + bk + c)$

Step 3(b): Show that a particular solution of the form yp=Axekx,whereA=1/(2ak+b)

If k is a root of this auxiliary equation with one multiplication, then we can have the homogeneous solution of our given differential equation as $y_{h} = c_{1} e^{k x} + c_{2} e^{m_{2} x}, w h e r e m_{1} = k$

Now we have to obtain the particular solution by assuming that $y_{p} = A x e^{k x}$ , where the function $g (x) = A e^{k x}$ and we timed it by x because it is $e^{k x}$ already described in the homogeneous solution showed above, and we have to obtain an additional solution as the following technique :

$y_{p}' = k A x e^{k x} + A e^{k x} . . . . (4) y_{p}'' = k^{2} A x e^{k x} + 2 K A e^{k x} . . . . (5)$

Now we have to substitute with $y_{p} = A x e^{k x}$ and equations (4) and (5) into equation (1), then we can have

$a (k^{2} A x e^{k x} + 2 K A e^{k x}) + b (k A x e^{k x} + A e^{k x}) + c A x e^{k x} = e^{k x} a k^{2} A x e^{k x} + 2 a K A e^{k x} + b k A x e^{k x} + b A e^{k x} + c A x e^{k x} = e^{k x} a k^{2} A + 2 a K A + b k A x + b A + c A x = 1 (a k^{2} A + b k A + c A) x + (2 a K A + b A) = 1$

Then we have

$2 a K A + b A = 1$

Then we obtain the equation

$A = \frac{1}{2 a k + b}$

Now using the equation $A = \frac{1}{2 a k + b}$

We can show that

$k \neq \frac{- b}{2 a} a s$

$A = \frac{1}{2 a \times \frac{- b}{2 a} + b} = \frac{1}{- b + b}, I t c a n n o t b e$

where the denominator must not equal zero.

Hence, it is shown that we can find a particular solution of the form a particular solution of the form localid="1663926985160" $y_{p} = A x e^{k x}, w h e r e A = 1 / (2 a k + b)$

Step 4(c): Show that a particular solution of the form y=Ax2ekx,whereA=1/(2a)

If is a root of this auxiliary equation with two multiplication, then we can have the homogeneous solution of our given differential equation as

$y_{h} = c_{1} e^{k x} + c_{2} x e^{k x}, w h e r e m_{1} = m_{2} = k$

Now we have to obtain the particular solution by assuming that $y_{p} = A x^{2} e^{k s}$ where the function $g (x) = A e^{k x}$ , and we timed it by $x^{2}$ because the part $x^{2} (e^{k x} + x e^{k x})$ is already described in the homogeneous solution showed above, and we have to obtain an additional solution as the following technique :

Differentiate the assumption with respect to x, then we have

$y_{p}' = k A x^{2} e^{k x} + 2 A x e^{k x} \cdot \cdot \cdot \cdot \cdot \cdot (6) y_{p}'' = k^{2} A x^{2} e^{k x} + 4 K A x e^{k x} + 2 A e^{k x} \cdot \cdot \cdot \cdot \cdot \cdot (7)$

Now we have to substitute with $y_{p} = A x^{2} e^{k s}$ and equations (6) and (7) into equation (1) , then we can have

$a (k^{2} A x^{2} e^{k x} + 4 K A x e^{k x} + 2 A e^{k x}) + b (k A x^{2} e^{k x} + 2 A x e^{k x}) + c A x^{2} e^{k x} = e^{k x} a k^{2} A x^{2} e^{k x} + 4 a K A x e^{k x} + 2 a A e^{k x} + b k A x^{2} e^{k x} + 2 b A x e^{k x} + c A x^{2} e^{k x} = e^{k x} a k^{2} A x^{2} + 4 a K A x + 2 a A + b k A x^{2} + 2 b A x + c A x^{2} = 1 (a k^{2} A + b k A + c A) x^{2} + (4 a k A + 2 b A + c A) x + (2 a A) = 1$

Then we have

$2 a A = 1$

Then we obtain the equation

$A = \frac{1}{2 a}$

Hence, it is shown that we can find a particular solution of the form a particular solution of the form $y = A x^{2} e^{k x}, w h e r e A = 1 / (2 a)$

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Short Answer

Step by step solution

Given information

Step 2(a): Show that a particular solution of the form yg=Aekx , where A=1/(ak2+bk+c)

Step 3(b): Show that a particular solution of the form yp=Axekx,whereA=1/(2ak+b)

Step 4(c): Show that a particular solution of the form y=Ax2ekx,whereA=1/(2a)

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