Question 51 of 96intermediate🔧 ApplyLong Answer4 marks

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

x2dydx=x2−2y2+xyx^2 \frac{dy}{dx} = x^2 - 2y^2 + xy
Correct Answer

The differential equation is homogeneous. The general solution is:

x+2yx−2y=C⋅x22\frac{x + \sqrt{2}y}{x - \sqrt{2}y} = C \cdot x^{2\sqrt{2}}
Exercise: EXERCISE 9.4 | Q: 5 | (Page 22)
For More Understanding

Explanation

This question tests the concept of homogeneous differential equations from Exercise 9.4. A homogeneous differential equation can be expressed in the form dydx=f ⁣(yx)\frac{dy}{dx} = f\!\left(\frac{y}{x}\right). The solution involves substituting y=vxy = vx to separate variables, then integrating both sides.

Solution Steps

  1. Step 1: Rewrite the given equation:

    x2dydx=x2−2y2+xyx^2 \frac{dy}{dx} = x^2 - 2y^2 + xy

  2. Step 2: Divide both sides by x2x^2:

    dydx=1−2(yx)2+yx\frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}

  3. Step 3: Since RHS is a function of yx\frac{y}{x} only, the equation is homogeneous.

  4. Step 4: Substitute y=vxy = vx, so:

    dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}

  5. Step 5: Substituting:

    v+xdvdx=1−2v2+vv + x\frac{dv}{dx} = 1 - 2v^2 + v

  6. Step 6: Simplify:

    xdvdx=1−2v2x\frac{dv}{dx} = 1 - 2v^2

  7. Step 7: Separate variables:

    dv1−2v2=dxx\frac{dv}{1 - 2v^2} = \frac{dx}{x}

  8. Step 8: Integrate both sides:

    122ln⁡∣1+2v1−2v∣=ln⁡∣x∣+C\frac{1}{2\sqrt{2}}\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = \ln|x| + C

  9. Step 9: Simplify:

    ln⁡∣1+2v1−2v∣=22ln⁡∣x∣+C′\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = 2\sqrt{2}\ln|x| + C'

  10. Step 10: Substitute back v=yxv = \frac{y}{x}:

    x+2yx−2y=C⋅x22\frac{x + \sqrt{2}y}{x - \sqrt{2}y} = C \cdot x^{2\sqrt{2}}
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