PROVING GIVEN RELATIONDHIP FOR THE DERIVATIVE OF EXPONENTIAL FUNCTION

Given the function on the left, demonstrate that the relationship is true.

Problem 1 :

For y = e^x + e^-x, prove that y’’ = y

Solution :

y = e^x + e^-x

First derivative:

y = e^x + e^-x

y' = e^x - e^-x

Second derivative:

y’’ = e^x - e^-x (-1)

y’’ = e^x + e^-x

Here e^x + e^-x = y

By applying the value, we get

y’’ = y

So, the given function is true.

Problem 2:

For y = 4e^-x + 5e^-3x, prove that y’’ + 4y’ + 3y = 0

Solution:

y = 4e^-x + 5e^-3x

First derivative:   

y’ = 4e^-x (-1) + 5e^-3x (-3)

y’ = -4e^-x - 15e^-3x

Second derivative:

y’’ = -4e^-x (-1) - 15e^-3x (-3)

y’’ = 4e^-x + 45e^-3x

From right side,

y’’ + 4y’ + 3y

= 4e^-x + 45e^-3x + 4(-4e^-x - 15e^-3x) + 3(4e^-x + 5e^-3x)

= 4e^-x + 45e^-3x - 16e^-x - 60e^-3x + 12e^-x + 15e^-3x

= 4e^-x - 16e^-x+ 12e^-x + 45e^-3x - 60e^-3x + 15e^-3x

= 0

So, the given function is true.

Problem 3 :

For y = e^2x + e^8x, prove that y’’ - 10y’ + 16y = 0

Solution :

y = e^2x + e^8x

First derivative :    

y' = 2e^2x + 8e^8x

Second derivative :

y’’ = 2e^2x (2) + 8e^8x(8)

y’’ = 4e^2x + 64e^8x

From right side,

y’’ - 10y’ + 16y

= 4e^2x + 64e^8x - 10(2e^2x + 8e^8x) + 16(e^2x + e^8x)

= 4e^2x - 20e^2x+ 16e^2x + 64e^8x + 16e^8x - 80e^8x  

y’’ - 10y’ + 16y = 0

So, the given function is true.

Problem 4 :

For y = e^2x + e^4x, prove that y’’ - 6y’ + 8y = 0

Solution :

y = e^2x + e^4x

First derivative :    

y' = e^2x(2) + e^4x(4)

y’ = 2e^2x + 4e^4x

Second derivative :

y’' = 2e^2x(2) + 4e^4x (4)

y’' = 4e^2x + 16e^4x

From right side,

y’’ - 6y’ + 8y = 4e^2x + 16e^4x - 6(2e^2x + 4e^4x) + 8(e^2x + e^4x)

= 4e^2x + 8e^2x - 12e^2x + 16e^4x + 8e^4x - 24e^4x

y’’ - 6y’ + 8y = 0

So, the given function is true.

Problem 5 :

For y = (x + 1) e^5x, prove that y’’ - 10y’ + 25y = 0

Solution :

y = (x + 1) e^5x

First derivative:    

Using product rule, we can find the derivative.

u = x + 1 and v = e^5x

u' = 1 and v' = 5e^5x

= (1) e^5x + (x + 1) e^5x (5)

y’ = e^5x + 5(x + 1) e^5x

= e^5x + 5xe^5x + 5e^5x

= 5xe^5x + 6e^5x

y’ = e^5x (5x + 6)

Second derivative:

u = 5x + 6 and v = e^5x

u' = 5 and v' = 5e^5x

y’’ = e^5x (5) + (5x + 6) e^5x (5)

= 5e^5x + 5e^5x (5x + 6)

= 5e^5x + 25xe^5x + 30e^5x

= 25xe^5x + 35e^5x

y’’ = e^5x (25x + 35)

From right side,

y’’ - 10y’ + 25y

= 25xe^5x + 35e^5x - 10 (5xe^5x + 6e^5x) + 25((x + 1) e^5x)

= 25xe^5x + 35e^5x - 50xe^5x - 60

e^5x + 25xe^5x + 25e^5x

y’’ - 10y’ + 25y = 0   

So, the given function is true.

Problem 6 :

For y = A + Be^-4x, prove that y’’ + 4y’ = 0

Solution :

y = A + Be^-4x

First derivative :

y' = A + Be^-4x (-4)

= (0) + Be^-4x (-4)

y’ = -4Be^-4x

Second derivative :

y’’ = -4 Be^-4x

= -4Be^-4x (-4)

y’’ = 16Be^-4x

From right side,

y’’ + 4y’ = 16Be^-4x + 4(-4Be^-4x)

= 16Be^-4x - 16Be^-4x

y’’ + 4y’ = 0

So, the given function is true.