Question 38 of 96intermediate🔧 ApplyLong Answer4 marks

Find the equation of a curve passing through the point (0,0)(0, 0) and whose differential equation is y′=exsin⁡xy' = e^x \sin x.

Correct Answer

y = ex(sin⁡x−cos⁡x)+12\frac{e^x(\sin x - \cos x) + 1}{2}

Exercise: EXERCISE 9.3 | Q: 15 | (Page 12)
For More Understanding

Explanation

This is a direct integration problem. The differential equation y′=exsin⁡xy' = e^x \sin x can be solved by integrating both sides with respect to xx. The integral of exsin⁡xe^x \sin x requires integration by parts applied twice.

After finding the general solution, the constant CC is determined using the condition that the curve passes through (0,0)(0, 0).

Solution Steps

  1. Step 1: Given differential equation is dydx=e\frac{dy}{dx} = e^x sin⁡\sin x. Integrating both sides: y = ∫\int exe^{x} sin⁡\sin x , dx + C

  2. Step 2: Using integration by parts for I = ∫\int exe^{x} sin⁡\sin x , dx: Let u = sin⁡\sin x, dv = exe^{x} dx. Then I = exe^{x} sin⁡\sin x - ∫\int exe^{x} cos⁡\cos x , dx

  3. Step 3: For ∫\int exe^{x} cos⁡\cos x , dx, apply integration by parts again: ∫\int exe^{x} cos⁡\cos x , dx = exe^{x} cos⁡\cos x + ∫\int exe^{x} sin⁡\sin x , dx = exe^{x} cos⁡\cos x + I

  4. Step 4: Substituting back: I = exe^{x} sin⁡\sin x - exe^{x} cos⁡\cos x - I, giving 2I = exe^{x}(sin⁡\sin x - cos⁡\cos x), so I = ex(sin⁡x−cos⁡x)2\frac{e^x(\sin x - \cos x)}{2}

  5. Step 5: Therefore, y = ex(sin⁡x−cos⁡x)2\frac{e^x(\sin x - \cos x)}{2} + C. Using point (0, 0): 0 = e0(sin⁡0−cos⁡0)2\frac{e^0(\sin 0 - \cos 0)}{2} + C = 1(0−1)2\frac{1(0 - 1)}{2} + C = -12\frac{1}{2} + C, so C = 12\frac{1}{2}

  6. Step 6: The equation of the curve is y = ex(sin⁡x−cos⁡x)+12\frac{e^x(\sin x - \cos x) + 1}{2}

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