Question

Consider the differential equation
$\cos(y) \frac{dy}{dx} + \frac{1}{x} \sin(y) = x, \quad (x > 0)$;
given that $y = \frac{\pi}{2}$ at $x = \sqrt{3}$. Which one of the following is the value of $y$ at $x = \sqrt{\frac{3}{2}}$ ?

• A. $\pi/6$
• B. $\pi/3$
• C. $\pi/2$
• D. $\pi/4$

Hint:

Notice that substituting \(u = \sin(y)\) simplifies the equation into a very common linear differential equation.
Always verify the value of the constant of integration \(C\) first to avoid carrying over algebraic errors.

Answer 1

The Correct Option is A

•

Step 1 : Understanding the Question:
We are given a first-order non-linear differential equation.
We are also given an initial condition: \(y = \frac{\pi}{2}\) when \(x = \sqrt{3}\).
We need to find the value of \(y\) at \(x = \sqrt{\frac{3}{2}}\).

•

Step 2 : Key Formula or Approach:
We can transform this non-linear differential equation into a linear one using substitution.
Let \(u = \sin(y)\)., then \(\frac{du}{dx} = \cos(y) \frac{dy}{dx}\).
This gives a first-order linear differential equation in \(u\) of the form:
\[ \frac{du}{dx} + P(x) u = Q(x) \] We solve this using the Integrating Factor method:
\[ \text{I.F.} = e^{\int P(x) dx} \] The solution is given by:
\[ u \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C \]
•

Step 3 : Detailed Explanation:
Let us apply the substitution:
\[ u = \sin(y) \implies \frac{du}{dx} = \cos(y) \frac{dy}{dx} \] Substituting this into the differential equation:
\[ \frac{du}{dx} + \frac{1}{x} u = x \] This is a linear first-order differential equation with \(P(x) = \frac{1}{x}\) and \(Q(x) = x\).
Let us find the Integrating Factor (I.F.):
\[ \text{I.F.} = e^{\int \frac{1}{x} dx} = e^{\log(x)} = x \] Multiplying both sides by the integrating factor:
\[ x \frac{du}{dx} + u = x^2 \implies \frac{d}{dx}(ux) = x^2 \] Integrating both sides with respect to \(x\):
\[ ux = \frac{x^3}{3} + C \] Substitute back \(u = \sin(y)\):
\[ x \sin(y) = \frac{x^3}{3} + C \] Now we apply the boundary condition: \(y = \frac{\pi}{2}\) when \(x = \sqrt{3}\).
\[ \sqrt{3} \sin\left(\frac{\pi}{2}\right) = \frac{(\sqrt{3})^3}{3} + C \] Since \(\sin\left(\frac{\pi}{2}\right) = 1\) and \((\sqrt{3})^3 = 3\sqrt{3}\):
\[ \sqrt{3} = \sqrt{3} + C \implies C = 0 \] So the solution is:
\[ x \sin(y) = \frac{x^3}{3} \implies \sin(y) = \frac{x^2}{3} \] Now, let us find the value of \(y\) at \(x = \sqrt{\frac{3}{2}}\):
\[ \sin(y) = \frac{\left(\sqrt{\frac{3}{2}}\right)^2}{3} = \frac{3/2}{3} = \frac{1}{2} \] Since \(\sin(y) = \frac{1}{2}\)., the principal value of \(y\) is:
\[ y = \frac{\pi}{6} \]
•

Step 4 : Final Answer:
The value of \(y\) at \(x = \sqrt{\frac{3}{2}}\) is \(\frac{\pi}{6}\).
This corresponds to option (A).