Question 60 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

dydx−yx+cos⁡sec⁡(yx)=0;y=0 when x=1\frac{dy}{dx} - \frac{y}{x} + \cos \sec\left(\frac{y}{x}\right) = 0; \quad y = 0 \text{ when } x = 1
Correct Answer

The particular solution is:

cos⁡(yx)−ln⁡∣x∣=1\cos\left(\frac{y}{x}\right) - \ln|x| = 1
Exercise: EXERCISE 9.4 | Q: 14 | (Page 22)
For More Understanding

Explanation

This is a homogeneous differential equation. The student needs to identify that the equation can be written in terms of y/xy/x, make the substitution y=vxy = vx, separate variables, integrate, and apply the initial condition y=0y = 0 when x=1x = 1.

Solution Steps

  1. Step 1: Rewrite the differential equation as dydx=yx−cosec⁡(yx)\frac{dy}{dx} = \frac{y}{x} - \cosec\left(\frac{y}{x}\right). This is a homogeneous equation since the RHS is a function of y/xy/x.

  2. Step 2: Substitute y=vxy = vx, so dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}. The equation becomes:

    v+xdvdx=v−cosec⁡(v)v + x\frac{dv}{dx} = v - \cosec(v)

  3. Step 3: Simplify to get xdvdx=−cosec⁡(v)x\frac{dv}{dx} = -\cosec(v), which gives:

    sin⁡(v) dv=−dxx\sin(v)\,dv = -\frac{dx}{x}

  4. Step 4: Integrate both sides:

    ∫sin⁡(v) dv=−∫dxx\int \sin(v)\,dv = -\int \frac{dx}{x}

    This gives −cos⁡(v)=−ln⁡∣x∣+C-\cos(v) = -\ln|x| + C, or:

    cos⁡(v)=ln⁡∣x∣−C\cos(v) = \ln|x| - C

  5. Step 5: Substitute back v=yxv = \frac{y}{x} to get the general solution:

    cos⁡(yx)=ln⁡∣x∣−C\cos\left(\frac{y}{x}\right) = \ln|x| - C

  6. Step 6: Apply the initial condition y=0y = 0 when x=1x = 1:

    cos⁡(0)=ln⁡(1)−C\cos(0) = \ln(1) - C

    which gives 1=0−C1 = 0 - C, so C=−1C = -1.

  7. Step 7: The particular solution is:

    cos⁡(yx)=ln⁡∣x∣+1\cos\left(\frac{y}{x}\right) = \ln|x| + 1

    or

    cos⁡(yx)−ln⁡∣x∣=1\cos\left(\frac{y}{x}\right) - \ln|x| = 1

    ∴ cos⁡(yx)−ln⁡∣x∣=1\cos\left(\frac{y}{x}\right) - \ln|x| = \mathbf{1}

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