Question 91 of 96intermediate🔧 ApplyLong Answer4 marks

Solve the differential equation:

[e−2xx−yx]dxdy=1(x≠0)\left[ \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1 \quad (x \neq 0)
Correct Answer

The solution to the differential equation is:

y=(2x+C)e−2xy = (2\sqrt{x} + C)e^{-2\sqrt{x}}
Exercise: Miscellaneous Exercise on Chapter 9 | Q: 10 | (Page 36)
For More Understanding

Explanation

The question asks to solve a differential equation listed in the provided context (Question 10). Although the context provided focuses on Separation of Variables and Homogeneous equations, Question 10 is a standard Linear Differential Equation. An expert teacher would recognize the form dydx+Py=Q\frac{dy}{dx} + Py = Q and apply the Integrating Factor method.

The steps involve algebraic rearrangement, finding the integrating factor e∫Pdxe^{\int P dx}, and integrating the resulting expression. The integration techniques used (∫1xdx\int \frac{1}{\sqrt{x}} dx and exponential rules) are consistent with the mathematical operations demonstrated in the provided context chunks (e.g., Example 20 and 6).

Solution Steps

  1. Step 1: Rearrange the given equation

    [e−2xx−yx]dxdy=1\left[ \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1

    into the form

    dydx=e−2x−yx=e−2xx−yx\frac{dy}{dx} = \frac{e^{-2\sqrt{x}} - y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}

  2. Step 2: Rewrite it as a linear differential equation:

    dydx+yx=e−2xx\frac{dy}{dx} + \frac{y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}

    Here, P=1xP = \frac{1}{\sqrt{x}} and Q=e−2xxQ = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}.

  3. Step 3: Calculate the Integrating Factor (IF):

    IF=e∫P dx=e∫1x dx=e2x\text{IF} = e^{\int P \, dx} = e^{\int \frac{1}{\sqrt{x}} \, dx} = e^{2\sqrt{x}}

  4. Step 4: Multiply the equation by the IF:

    y e2x=∫(e−2xx⋅e2x)dx+Cy \, e^{2\sqrt{x}} = \int \left( \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \cdot e^{2\sqrt{x}} \right) dx + C

  5. Step 5: Simplify and integrate the right side:

    y e2x=∫1x dx+C=2x+Cy \, e^{2\sqrt{x}} = \int \frac{1}{\sqrt{x}} \, dx + C = 2\sqrt{x} + C

  6. Step 6: Isolate yy to get the final solution:

    y=(2x+C) e−2xy = (2\sqrt{x} + C) \, e^{-2\sqrt{x}}

    ∴ The solution is y=(2x+C) e−2x\mathbf{y = (2\sqrt{x} + C) \, e^{-2\sqrt{x}}}.

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