y^(')=0.5(100-y) Is y=5e^(-0.5 x)+100 a solution to the above equation? Choose 1 answer: (A) Yes (B) No

Given Differential Equation: Given the differential equation y'=0.5(100-y), we need to verify if the function y=5e^(-0.5x)+100 is a solution to this equation.Differentiation of y: To verify if y=5e^(-0.5x)+100 is a solution, we need to differentiate this function with respect to x and check if the derivative matches the right-hand side of the given differential equation.Substitute into Equation: Differentiate y=5e^(-0.5x)+100 with respect to x using the chain rule for the exponential function. y' = d/dx [5e^(-0.5x)] + d/dx [100] y' = 5 * (-0.5) * e^(-0.5x) + 0 y' = -2.5e^(-0.5x)Simplify Expression: Now, we substitute y=5e^(-0.5x)+100 into the right-hand side of the given differential equation to see if it matches the derivative we found. Right-hand side of the differential equation: 0.5(100-y) = 0.5(100-(5e^(-0.5x)+100))Compare Derivative: Simplify the right-hand side expression. 0.5(100-(5e^(-0.5x)+100)) = 0.5(100-5e^(-0.5x)-100) 0.5(-5e^(-0.5x)) = -2.5e^(-0.5x)Compare the derivative y' with the simplified right-hand side of the differential equation. We found y' = -2.5e^(-0.5x) and the right-hand side is also -2.5e^(-0.5x). Since both expressions are equal, y=5e^(-0.5x)+100 is indeed a solution to the differential equation y'=0.5(100-y).

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y^(')=0.5(100-y)
Is
y=5e^(-0.5 x)+100 a solution to the above equation?
Choose 1 answer:
(A) Yes
(B) No

$y^{\prime}=0.5(100-y)$ $\newline$ Is $y=5 e^{-0.5 x}+100$ a solution to the above equation? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) No

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Full solution

y^{\prime}=0.5(100-y)

\newline

y=5 e^{-0.5 x}+100

a solution to the above equation?

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Given Differential Equation: Given the differential equation $y'=0.5(100-y)$ , we need to verify if the function $y=5e^{-0.5x}+100$ is a solution to this equation.
Differentiation of $y$ : To verify if $y=5e^{-0.5x}+100$ is a solution, we need to differentiate this function with respect to $x$ and check if the derivative matches the right-hand side of the given differential equation.
Substitute into Equation: Differentiate $y=5e^{-0.5x}+100$ with respect to $x$ using the chain rule for the exponential function. $\newline$ $y' = \frac{d}{dx} [5e^{-0.5x}] + \frac{d}{dx} [100]$ $\newline$ $y' = 5 * (-0.5) * e^{-0.5x} + 0$ $\newline$ $y' = -2.5e^{-0.5x}$
Simplify Expression: Now, we substitute $y=5e^{-0.5x}+100$ into the right-hand side of the given differential equation to see if it matches the derivative we found. $\newline$ Right-hand side of the differential equation: $0.5(100-y) = 0.5(100-(5e^{-0.5x}+100))$
Compare Derivative: Simplify the right-hand side expression. $\newline$ $0.5(100-(5e^{-0.5x}+100)) = 0.5(100-5e^{-0.5x}-100)$ $\newline$ $0.5(-5e^{-0.5x}) = -2.5e^{-0.5x}$
Compare Derivative: Simplify the right-hand side expression. $\newline$ $0.5(100-(5e^{-0.5x}+100)) = 0.5(100-5e^{-0.5x}-100)$ $\newline$ $0.5(-5e^{-0.5x}) = -2.5e^{-0.5x}$ Compare the derivative $y'$ with the simplified right-hand side of the differential equation. $\newline$ We found $y' = -2.5e^{-0.5x}$ and the right-hand side is also $-2.5e^{-0.5x}$ . $\newline$ Since both expressions are equal, $y=5e^{-0.5x}+100$ is indeed a solution to the differential equation $y'=0.5(100-y)$ .