If (dy)/(dx)=2-y, and if y=1 when x=1, then y= (A) 2-e^(x-1) (B) 2-e^(1-x) (C) 2-e^(-x) (D) 2+e^(-x)

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Write Differential Equation: Write down the given differential equation and the initial condition. We have the differential equation (dy)/(dx) = 2 - y and the initial condition y(1) = 1.Separate Variables: Recognize that the differential equation is a first-order linear ordinary differential equation that can be solved using separation of variables. We can rearrange the equation to separate the variables y and x: dy/(2 - y) = dxIntegrate Equations: Integrate both sides of the equation with respect to their respective variables. ∫(1/(2 - y)) dy = ∫dxCombine Constants: Perform the integration on both sides. The left side requires a substitution to integrate. Let u = 2 - y, then du = -dy. The integral becomes -∫(1/u) du. The right side is straightforward: ∫dx = x + C, where C is the constant of integration. -∫(1/u) du = -ln|u| + C1 x + C = -ln|2 - y| + C1Solve for y: Combine the constants of integration into a single constant. Since C1 is also a constant, we can write -ln|2 - y| + C1 as -ln|2 - y| + C2, where C2 is a new constant that combines C and C1. x + C2 = -ln|2 - y|Use Initial Condition: Solve for y in terms of x. To isolate y, we first exponentiate both sides to remove the natural logarithm: e^(x + C2) = e^(-ln|2 - y|) e^(x + C2) = 1/(2 - y) Now, solve for y: 2 - y = 1/e^(x + C2) y = 2 - 1/e^(x + C2)Find Constant C2: Use the initial condition to find the value of the constant C2. We know that y(1) = 1, so we substitute x = 1 and y = 1 into the equation: 1 = 2 - 1/e^(1 + C2) Solve for e^(1 + C2): e^(1 + C2) = 2 - 1 e^(1 + C2) = 1 Now, take the natural logarithm of both sides to solve for C2: 1 + C2 = ln(1) C2 = ln(1) - 1 C2 = 0 - 1 C2 = -1Substitute C2 into y: Substitute the value of C2 back into the equation for y. y = 2 - 1/e^(x - 1) This matches answer choice (A) 2 - e^(x - 1).

If $\frac{d y}{d x}=2-y$ , and if $y=1$ when $x=1$ , then $y=$ $\newline$ (A) $2-e^{x-1}$ $\newline$ (B) $2-e^{1-x}$ $\newline$ (C) $2-e^{-x}$ $\newline$ (D) $2+e^{-x}$

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If dydx=2−y \frac{d y}{d x}=2-y dxdy​=2−y, and if y=1 y=1 y=1 when x=1 x=1 x=1, then y= y= y=\newline(A) 2−ex−1 2-e^{x-1} 2−ex−1\newline(B) 2−e1−x 2-e^{1-x} 2−e1−x\newline(C) 2−e−x 2-e^{-x} 2−e−x\newline(D) 2+e−x 2+e^{-x} 2+e−x

If $\frac{d y}{d x}=2-y$ , and if $y=1$ when $x=1$ , then $y=$ $\newline$ (A) $2-e^{x-1}$ $\newline$ (B) $2-e^{1-x}$ $\newline$ (C) $2-e^{-x}$ $\newline$ (D) $2+e^{-x}$