Question

The Solution of the differential equation (D - 1)2y = 0 is......​

Answer 1

The solution of the differential equation is given by

$\displaystyle \sf{ y = (a + bx) {e}^{x} }$

Given :

The differential equation (D – 1)² y = 0

To find :

The solution of the equation

Solution :

Step 1 of 3 :

Write down the given differential equation

The given differential equation is

(D – 1)² y = 0

Step 2 of 3 :

Find the auxiliary equation

Let $\displaystyle \sf{ y = {e}^{mx} }$be the trial solution

The auxiliary equation is given by

(m – 1)² = 0

Step 3 of 3 :

Find solution of the differential equation

(m – 1)² = 0

⇒ m = 1 , 1

So the roots are real and equal

Hence the required solution is given by

$\displaystyle \sf{ y = (a + bx) {e}^{x} }$

Where a and b are constants

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Answer 2

Answer:

The solution of the differential equation is given by

\displaystyle \sf{ y = (a + bx) {e}^{x} }y=(a+bx)e

x

Given :

The differential equation (D – 1)² y = 0

To find :

The solution of the equation

Solution :

Step 1 of 3 :

Write down the given differential equation

The given differential equation is

(D – 1)² y = 0

Step 2 of 3 :

Find the auxiliary equation

Let \displaystyle \sf{ y = {e}^{mx} }y=e

mx

be the trial solution

The auxiliary equation is given by

(m – 1)² = 0

Step 3 of 3 :

Find solution of the differential equation

(m – 1)² = 0

⇒ m = 1 , 1

So the roots are real and equal

Hence the required solution is given by

\displaystyle \sf{ y = (a + bx) {e}^{x} }y=(a+bx)e

x

Step-by-step explanation:

The solution of the differential equation is given by

\displaystyle \sf{ y = (a + bx) {e}^{x} }y=(a+bx)e

x

Given :

The differential equation (D – 1)² y = 0

To find :

The solution of the equation

Solution :

Step 1 of 3 :

Write down the given differential equation

The given differential equation is

(D – 1)² y = 0

Step 2 of 3 :

Find the auxiliary equation

Let \displaystyle \sf{ y = {e}^{mx} }y=e

mx

be the trial solution

The auxiliary equation is given by

(m – 1)² = 0

Step 3 of 3 :

Find solution of the differential equation

(m – 1)² = 0

⇒ m = 1 , 1

So the roots are real and equal

Hence the required solution is given by

\displaystyle \sf{ y = (a + bx) {e}^{x} }y=(a+bx)e

x