Question 65 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations given in Exercises 1 to 12, find the general solution:

dydx+3y=e−2x\frac{dy}{dx} + 3y = e^{-2x}
Correct Answer

The general solution is:

y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x}
Exercise: EXERCISE 9.5 | Q: 2 | (Page 29)
For More Understanding

Explanation

This is a linear differential equation of the form dydx+Py=Q\frac{dy}{dx} + Py = Q. The student needs to identify P=3P = 3 and Q=e−2xQ = e^{-2x}, then find the integrating factor and integrate.

The context shows similar methodology in the worked example where the integrating factor method is applied.

Solution Steps

  1. Step 1: Identify the form of the differential equation.

    The given equation

    dydx+3y=e−2x\frac{dy}{dx} + 3y = e^{-2x}

    is of the form

    dydx+Py=Q\frac{dy}{dx} + Py = Q

    where P=3P = 3 and Q=e−2xQ = e^{-2x}.

  2. Step 2: Find the Integrating Factor (I.F.).

    I.F.=e∫P dx=e∫3 dx=e3xI.F. = e^{\int P \, dx} = e^{\int 3 \, dx} = e^{3x}

  3. Step 3: Apply the formula for solution.

    The solution is given by:

    y×(I.F.)=∫Q×(I.F.) dx+Cy \times (I.F.) = \int Q \times (I.F.) \, dx + C

    Substituting the values:

    y⋅e3x=∫e−2x⋅e3x dx+Cy \cdot e^{3x} = \int e^{-2x} \cdot e^{3x} \, dx + C

  4. Step 4: Simplify and integrate.

    y⋅e3x=∫ex dx+Cy \cdot e^{3x} = \int e^{x} \, dx + C y⋅e3x=ex+Cy \cdot e^{3x} = e^{x} + C

  5. Step 5: Solve for y.

    Dividing both sides by e3xe^{3x}:

    y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x}

    This is the required general solution.

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