Question

Check whether each of the following functions is a solution of the differential equation 2y' + 5y = 3e^-x
(a) y = e^-x
2y' = ___
5y = ___
2y' + 5y = ___
(b) e^-x + e^-(5/2)x
2y' = ___
5y = ___
2y' + 5y = ___
(c) e^-x + Ce^-(5/2)x
2y' = ___
5y = ___
2y' + 5y = ___

Answer 1

To check whether a function is a solution of the given differential equation, we need to substitute the function into the equation and see if it satisfies the equation.  
Let's go through each function:  (a) y = e^-x  To find y', we take the derivative of y with respect to x: y' = -e^-x  Now let's substitute y' and y into the equation: 2y' = 2(-e^-x) = -2e^-x 5y = 5(e^-x) = 5e^-x 2y' + 5y = -2e^-x + 5e^-x = 3e^-x  Since 2y' + 5y equals 3e^-x, the function y = e^-x is a solution to the differential equation.  (b) y = e^-x + e^-(5/2)x  Let's find y' first: y' = -e^-x - (5/2)e^-(5/2)x  Substituting y' and y into the equation: 2y' = 2(-e^-x - (5/2)e^-(5/2)x) = -2e^-x - 5e^-(5/2)x 5y = 5(e^-x + e^-(5/2)x) = 5e^-x + 5e^-(5/2)x 2y' + 5y = -2e^-x - 5e^-(5/2)x + 5e^-x + 5e^-(5/2)x = 3e^-x  Again, 2y' + 5y equals 3e^-x. So, the function y = e^-x + e^-(5/2)x is a solution to the differential equation.  (c) y = e^-x + Ce^-(5/2)x (C is a constant)  Taking the derivative of y: y' = -e^-x - (5/2)Ce^-(5/2)x  Substituting y' and y into the equation: 2y' = 2(-e^-x - (5/2)Ce^-(5/2)x) = -2e^-x - 5Ce^-(5/2)x 5y = 5(e^-x + Ce^-(5/2)x) = 5e^-x + 5Ce^-(5/2)x 2y' + 5y = -2e^-x - 5Ce^-(5/2)x + 5e^-x + 5Ce^-(5/2)x = 3e^-x  Once again, 2y' + 5y equals 3e^-x. Therefore, the function y = e^-x + Ce^-(5/2)x (where C is a constant) is also a solution to the differential equation.  In conclusion, all three functions y = e^-x, y = e^-x + e^-(5/2)x, and y = e^-x + Ce^-(5/2)x (C is a constant) are solutions to the differential equation 2y' + 5y = 3e^-x.

Answer 2

Answer:

Refer below.

Step-by-step explanation:

The task involves checking whether the given functions are a solution to the given differential equation.

Given differential equation:

$2y'+5y=3e^{-x}$

Asummed solutions:

$\text{(a) } y = e^{-x}\\\\\text{(b) } y = e^{-x}+e^{-\frac{5}{2}x}\\\\\text{(c) } y = e^{-x}+Ce^{-\frac{5}{2}x}$

$\hrulefill$

Answering Part (a):$\hrulefill$

For the function y = e⁻ˣ we will find the first and second derivatives of 'y':

• y = e⁻ˣ
• y' = -e⁻ˣ

Substitute these expressions into the differential equation:

$\Longrightarrow 2\left(-e^{-x}\right)+5\left(e^{-x}\right)=3e^{-x}$

Simplify the above equation, if the L.H.S equals the R.H.S then 'y' is a solution to the differential equation:

$\Longrightarrow -2e^{-x}+5e^{-x}=3e^{-x}\\\\\\\\\therefore 3e^{-x}=3e^{-x} \ \checkmark$

Thus, y = e⁻ˣ is a solution to the given differential equation.

$\hrulefill$

Answering Part (b):$\hrulefill$

For the function y = e⁻ˣ + e^(-5x/2) we will find the first and second derivatives of 'y':

$\bullet \ y = e^{-x}+e^{-\frac{5}{2}x}\\\\\bullet \ y' = -e^{-x}-\dfrac{5}{2}e^{-\frac{5}{2}x}$

Substitute these expressions into the differential equation:

$\Longrightarrow 2\left( -e^{-x}-\dfrac{5}{2}e^{-\frac{5}{2}x}\right)+5\left( e^{-x}+e^{-\frac{5}{2}x}\right)=3e^{-x}$

Simplify the above equation, if the L.H.S equals the R.H.S then 'y' is a solution to the differential equation:

$\Longrightarrow -2e^{-x}-5e^{-\frac{5}{2}x}+ 5e^{-x}+5e^{-\frac{5}{2}x}=3e^{-x}$

$\therefore 3e^{-x}=3e^{-x} \ \checkmark$

Thus, y = e⁻ˣ + e^(-5x/2) is a solution to the given differential equation.
$\hrulefill$

Answering Part (c):$\hrulefill$
For the function y = e⁻ˣ + Ce^(-5x/2) we will find the first and second derivatives of 'y':

$\bullet \ y = e^{-x}+Ce^{-\frac{5}{2}x}\\\\\bullet \ y' = -e^{-x}-\bar Ce^{-\frac{5}{2}x}$

Substitute these expressions into the differential equation:

$\Longrightarrow 2\left( -e^{-x}-\bar Ce^{-\frac{5}{2}x}\right)+5\left( e^{-x}+Ce^{-\frac{5}{2}x}\right)=3e^{-x}$

Simplify the above equation, if the L.H.S equals the R.H.S then 'y' is a solution to the differential equation:

$\Longrightarrow -2e^{-x}- \bar Ce^{-\frac{5}{2}x}\right)+5e^{-x}+\bar Ce^{-\frac{5}{2}x}\right)=3e^{-x}$

$\therefore 3e^{-x}+Ce^{-\frac{5}{2}x}=3e^{-x}$

Thus, y = e⁻ˣ + Ce^(-5x/2) may or may not be a solution to the DE, it will depend upon the value of the arbitrary constant 'C'.