Question 27 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations in Exercises 1 to 10, find the general solution:

dydx+y=1(y≠1)\frac{dy}{dx} + y = 1 \quad (y \neq 1)
Correct Answer

y = 1 + Ce−xCe^{-x}

Exercise: EXERCISE 9.3 | Q: 3 | (Page 12)
For More Understanding

Explanation

This is a first-order linear differential equation of the form dydx+Py=Q\frac{dy}{dx} + Py = Q. The solution requires finding the integrating factor and then integrating both sides.

Solution Steps

  1. Step 1: The given differential equation is dydx+y=1\frac{dy}{dx} + y = 1, which is of the form dydx+Py=Q\frac{dy}{dx} + Py = Q where P=1P = 1 and Q=1Q = 1.

  2. Step 2: Find the integrating factor (IF):

    IF=e∫P dx=e∫1 dx=ex\text{IF} = e^{\int P\, dx} = e^{\int 1\, dx} = e^x

  3. Step 3: Multiply both sides of the equation by the integrating factor exe^x:

    exdydx+ex⋅y=exe^x \frac{dy}{dx} + e^x \cdot y = e^x

  4. Step 4: The left side becomes ddx(yex)\frac{d}{dx}(ye^x). So,

    ddx(yex)=ex\frac{d}{dx}(ye^x) = e^x

  5. Step 5: Integrate both sides with respect to xx:

    yex=ex+Cye^x = e^x + C

  6. Step 6: Divide both sides by exe^x:

    y=1+Ce−xy = 1 + Ce^{-x}

  7. Step 7: Therefore, the general solution is:

    y=1+Ce−x\boxed{y = 1 + Ce^{-x}}

    where CC is an arbitrary constant.

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