Question 29 of 96intermediate🔧 ApplyLong Answer4 marks

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

(ex+e−x) dy−(ex−e−x) dx=0(e^x + e^{-x})\, dy - (e^x - e^{-x})\, dx = 0
Correct Answer

y = ln(exe^{x} + e−xe^{-x}) + C

Exercise: EXERCISE 9.3 | Q: 5 | (Page 12)
For More Understanding

Explanation

This is a separable differential equation from Exercise 9.4. The equation can be rearranged to separate variables xx and yy, then integrated using substitution method.

The key insight is recognizing that the numerator (ex−e−x)(e^x - e^{-x}) is the derivative of the denominator (ex+e−x)(e^x + e^{-x}), making substitution straightforward.

Solution Steps

  1. Step 1: Rearrange the given differential equation to separate variables.

    (ex+e−x) dy=(ex−e−x) dx(e^x + e^{-x})\, dy = (e^x - e^{-x})\, dx

  2. Step 2: Express in derivative form.

    dydx=ex−e−xex+e−x\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^x + e^{-x}}

  3. Step 3: Integrate both sides with respect to xx.

    ∫dy=∫ex−e−xex+e−x dx\int dy = \int \frac{e^x - e^{-x}}{e^x + e^{-x}}\, dx

  4. Step 4: Apply substitution method.

    Let t=ex+e−xt = e^x + e^{-x}

    Then dtdx=ex−e−x\dfrac{dt}{dx} = e^x - e^{-x}

    So dt=(ex−e−x) dxdt = (e^x - e^{-x})\, dx

  5. Step 5: Substitute and integrate.

    ∫ex−e−xex+e−x dx=∫dtt=ln⁡∣t∣+C\int \frac{e^x - e^{-x}}{e^x + e^{-x}}\, dx = \int \frac{dt}{t} = \ln|t| + C

    Since ex+e−x>0e^x + e^{-x} > 0 for all real xx, we have ln⁡(ex+e−x)+C\ln(e^x + e^{-x}) + C

  6. Step 6: Write the general solution.

    ∴ y=ln⁡(ex+e−x)+Cy = \ln(e^x + e^{-x}) + C

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