Question

find the complimentary and particular integral of (D^2+5D+6)y)=e^x​

Answer 1

Answer:

Let us first solve the homogeneous equation:

(D2−5D+6)yh=0

The homogeneous solution is:

(D−2)(D−3)yh=0⇒yh=Ae2x+Be3x⇒

⇒e−3xyh=Ae−x+B

Thus,

D(D+1)(e−3xyh)=0

Now, let us find a particlar solution for:

(D2+D)(e−3xyp)=x2

Take e−3xyp=a0+a1x+a2x2+a3x3

Then,

(e−3xyp)′=a1+2a2x+3a3x2⇒

⇒(e−3xyp)′′=2a2+6a3x⇒

⇒a1+2a2+(2a2+6a3)x+3a3x2=x2

So, we get that:

3a3=1⇒a3=13

2a2+6a3=0⇒2a2+2=0⇒a2=−1

a1+2a2=0⇒a1−2=0⇒a1=2

Thus,

e−3xyp=x33−x2+2x+C⇒

⇒yp=e3x(x33−x2+2x)+Ce3x⇒

⇒y=yh+yp=Ae2x+(x33−x2+2x+B)e3x

Answer 2

$\green{\large\underline{\sf{Solution-}}}$

Given Differential equation is

$\rm :\longmapsto\:( {D}^{2} + 5D + 6)y = {e}^{x}$

Its Auxiliary equation is

$\rm :\longmapsto\: {D}^{2} + 5D + 6 = 0$

$\rm :\longmapsto\: {D}^{2} + 3D + 2D + 6 = 0$

$\rm :\longmapsto\:D(D + 3) + 2(D + 3) = 0$

$\rm :\longmapsto\:(D + 3)(D + 2) = 0$

$\bf\implies \:D = - \: 2 \: \: or \: \: D = - \: 3$

So, Complementary function is

$\rm \implies\:\boxed{ \tt{ \: CF = c_1 {e}^{ - 2x} + c_2 {e}^{ - 3x} \: }}$

Now, Evaluation of Particular Integral

$\rm :\longmapsto\:PI = \dfrac{1}{ {D}^{2} + 5D + 6} {e}^{x}$

On substituting, D = 1, we get

$\rm :\longmapsto\:PI = \dfrac{1}{ 1 + 5 + 6} {e}^{x}$

$\rm \implies\:\boxed{ \tt{ \: PI = \frac{ {e}^{x} }{12} \: }}$

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Additional Information

Complete Solution of

$\rm :\longmapsto\:( {D}^{2} + 5D + 6)y = {e}^{x}$

$\rm \: = \: CF + PI$

$\rm \: = \: c_1 {e}^{ - 2x} + c_2 {e}^{ - 3x} + \dfrac{ {e}^{x} }{12}$