Question 78 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition:

dydx−3ycot⁡x=sin⁡2x;y=2 when x=π2\frac{dy}{dx} - 3y \cot x = \sin 2x; \quad y = 2 \text{ when } x = \frac{\pi}{2}
Correct Answer

y = 4 sin³x - 2 sin²x

Exercise: EXERCISE 9.5 | Q: 15 | (Page 30)
For More Understanding

Explanation

This is a linear differential equation of the form dydx+Py=Q\frac{dy}{dx} + Py = Q. The question asks for a particular solution satisfying the given condition, which is a standard method taught in NCERT using the integrating factor technique.

Solution Steps

  1. Step 1: Identify the given differential equation as linear form dydx+Py=Q\frac{dy}{dx} + Py = Q, where P=−3cot⁡xP = -3 \cot x and Q=sin⁡2xQ = \sin 2x.

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

    I.F.=e∫P dx=e∫(−3cot⁡x) dx=e−3log⁡∣sin⁡x∣=sin⁡−3x=1sin⁡3x\text{I.F.} = e^{\int P \, dx} = e^{\int (-3 \cot x) \, dx} = e^{-3 \log|\sin x|} = \sin^{-3}x = \frac{1}{\sin^3 x}

  3. Step 3: Apply the general solution formula:

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

  4. Step 4: Substitute values:

    ysin⁡3x=∫(sin⁡2x×1sin⁡3x)dx+C\frac{y}{\sin^3 x} = \int \left(\sin 2x \times \frac{1}{\sin^3 x}\right) dx + C

  5. Step 5: Simplify using sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x:

    ysin⁡3x=∫2cos⁡xsin⁡2x dx+C\frac{y}{\sin^3 x} = \int \frac{2 \cos x}{\sin^2 x} \, dx + C

  6. Step 6: Integrate:

    ysin⁡3x=−2sin⁡x+C\frac{y}{\sin^3 x} = \frac{-2}{\sin x} + C

  7. Step 7: Apply the condition y=2y = 2 when x=π2x = \frac{\pi}{2} (where sin⁡x=1\sin x = 1):

    21=−21+C\frac{2}{1} = \frac{-2}{1} + C

    giving C=4C = 4.

  8. Step 8: Substitute C=4C = 4 and solve for yy:

    y=4sin⁡3x−2sin⁡2xy = 4 \sin^3 x - 2 \sin^2 x

    ∴ y=4sin⁡3x−2sin⁡2x\mathbf{y = 4 \sin^3 x - 2 \sin^2 x}

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