Question 16 If d yd x 2 y and if y 1 when x 1 then y A 2 ex 1

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Question 16 If $\frac{d y}{d x}=2-y$, and if $y=1$ when $x=1$, then $y=$ (A) $2-e^{x-1}$

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#### Solution By Steps ***Step 1: Initial Condition*** Given $\frac{dy}{dx} = 2 - y$ and $y = 1$ when $x = 1$, we need to find the value of $y$. ***Step 2: Solve the Differential Equation*** Separate variables and solve the differential equation: $\frac{dy}{2-y} = dx$ Integrate both sides: $\int \frac{dy}{2-y} = \int dx$ ***Step 3: Integral Calculation*** Integrate to get: $- \ln|2-y| = x + C$ ***Step 4: Apply Initial Condition*** Use the initial condition $y = 1$ when $x = 1$ to find the constant $C$. ***Step 5: Find the Value of y*** Substitute $y = 1$ and $x = 1$ into the equation and solve for $C$. ***Step 6: Finalize the Solution*** Substitute the value of $C$ back into the equation to find the value of $y$. #### Final Answer $y = 2 - e^{1-x}$ #### Key Concept Differential Equations #### Key Concept Explanation Differential equations involve equations with derivatives. Solving them often requires integrating both sides and applying initial conditions to find the constants. They are crucial in modeling various real-world phenomena in physics, engineering, and many other fields.

Follow-up Knowledge or Question

What is the general solution to the differential equation $\frac{dy}{dx} = 2 - y$?

How can we use the initial condition $y=1$ when $x=1$ to find the particular solution to the differential equation?

What is the significance of the constant $e$ in the expression $2 - e^{x-1}$ obtained as the solution to the differential equation?

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If d yd x=2-y and if y=1 when x=1 then y= (A) 2-ex-1 (B) 2-e1-x (C) 2-e-x (D) 2+e-x

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Solve the differential equation x(y-1) d x+(x+1) d y=0 If y=2 when x=1 determine y when x=388

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If d yd x 2 y and if y 1 when x 1 then y A 2 ex 1 B 2 e1 x C 2 e x D 2 e x

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