Question:medium

The solution of the differential equation \[ \sqrt{a+x}\,\frac{dy}{dx} + xy = 0 \] is

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Use substitution \(t=a+x\) whenever \(\sqrt{a+x}\) appears—it simplifies integration instantly.
Updated On: May 21, 2026
  • \(y = Ce^{\frac{2}{3}(2a-x)\sqrt{x+a}}\)
  • \(y = Ce^{\frac{2}{3}(a-x)\sqrt{x+a}}\)
  • \(y = Ce^{\frac{2}{3}(2a+x)\sqrt{x+a}}\)
  • \(y = Ce^{-\frac{2}{3}(2a-x)\sqrt{x+a}}\)
Show Solution

The Correct Option is A

Solution and Explanation

The given differential equation is:

\[\sqrt{a + x} \frac{dy}{dx} + xy = 0\]

To solve this differential equation, we will use the integrating factor method. First, we rewrite the equation in the standard linear differential equation form:

\[\frac{dy}{dx} + \frac{x}{\sqrt{a + x}} y = 0\]

The standard form of a first-order linear differential equation is:

\[\frac{dy}{dx} + P(x)y = Q(x)\]

Here, \(P(x) = \frac{x}{\sqrt{a + x}}\) and \(Q(x) = 0\).

The integrating factor, \(\mu(x)\), is given by:

\[\mu(x) = e^{\int P(x) \, dx}\]

Substitute \(P(x)\) into the integrating factor formula:

\[\mu(x) = e^{\int \frac{x}{\sqrt{a + x}} \, dx}\]

Now, let's solve the integral:

\[\int \frac{x}{\sqrt{a + x}} \, dx\]

Let \(u = \sqrt{a + x}\). Then, \(x = u^2 - a\) and \(dx = 2u \, du\).

Substitute these into the integral:

\[\int \frac{u^2 - a}{u} \cdot 2u \, du = 2 \int (u^2 - a) \, du = 2 \left(\frac{u^3}{3} - au\right)+ C\]

We get:

\[\int \frac{x}{\sqrt{a + x}} \, dx = \frac{2}{3}(a + x)^{3/2} - 2a (a + x)^{1/2}+ C\]

Return to the integrating factor:

\[\mu(x) = e^{\frac{2}{3}(a + x)^{3/2} - 2a (a + x)^{1/2}}\]

Using the integrating factor, the solution to the differential equation is given by:

\[y \cdot \mu(x) = C\]

So, the solution is:

\(y = Ce^{\frac{2}{3}(2a-x)\sqrt{x+a}}\)

This matches the given correct answer. Thus, the solution to the differential equation is:

\(y = Ce^{\frac{2}{3}(2a-x)\sqrt{x+a}}\)

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