Question

If the solution of \(\dfrac{dy}{dx}=\dfrac{y^3\cos\sqrt{x}}{\sqrt{x}e^{1/y^2}}\), \(y(0)=1\), is \(\dfrac{1}{y^2}=\log_e(f(x))\), then \(f(x)=\)

• A. \(4+4\sin\sqrt{x}\)
• B. \(e\sin\sqrt{x}\)
• C. \(1-4\sin\sqrt{x}\)
• D. \(e-4\sin\sqrt{x}\)

Hint:

When a differential equation contains \(e^{1/y^2}\), try the substitution \[ u=\frac{1}{y^2}. \] This often converts the equation into a simpler separable form.

Answer 1

The Correct Option is D

Step 1: Rewrite the differential equation.
Given, \[ \frac{dy}{dx}=\frac{y^3\cos\sqrt{x}}{\sqrt{x}e^{1/y^2}} \] This can be written as \[ \frac{dy}{dx}=\frac{y^3\cos\sqrt{x}}{\sqrt{x}}e^{-1/y^2} \]

Step 2: Put \(u=\dfrac{1}{y^2}\).
Let \[ u=\frac{1}{y^2} \] Then, \[ \frac{du}{dx}=-\frac{2}{y^3}\frac{dy}{dx} \] Substituting \(\dfrac{dy}{dx}\), \[ \frac{du}{dx} = -\frac{2}{y^3}\cdot \frac{y^3\cos\sqrt{x}}{\sqrt{x}}e^{-u} \] \[ \frac{du}{dx} = -\frac{2\cos\sqrt{x}}{\sqrt{x}}e^{-u} \]

Step 3: Separate and integrate.
Multiplying by \(e^u\), \[ e^u\frac{du}{dx}=-\frac{2\cos\sqrt{x}}{\sqrt{x}} \] Since \[ \frac{d}{dx}(e^u)=e^u\frac{du}{dx}, \] we get \[ \frac{d}{dx}(e^u)=-\frac{2\cos\sqrt{x}}{\sqrt{x}} \] Integrating, \[ e^u=\int -\frac{2\cos\sqrt{x}}{\sqrt{x}}\,dx \] Put \[ t=\sqrt{x} \] Then, \[ dx=2t\,dt \] So, \[ e^u=\int -\frac{2\cos t}{t}\cdot 2t\,dt \] \[ e^u=\int -4\cos t\,dt \] \[ e^u=-4\sin t+C \] Thus, \[ e^u=-4\sin\sqrt{x}+C \]

Step 4: Use the initial condition.
Given, \[ y(0)=1 \] So, \[ u=\frac{1}{y^2}=1 \] At \(x=0\), \[ e^u=e^1=e \] Also, \[ \sin\sqrt{0}=0 \] Hence, \[ e=C \] Therefore, \[ e^u=e-4\sin\sqrt{x} \]

Step 5: Compare with the given form.
Since \[ u=\frac{1}{y^2}, \] we have \[ e^{1/y^2}=e-4\sin\sqrt{x} \] Taking logarithm, \[ \frac{1}{y^2}=\log_e(e-4\sin\sqrt{x}) \] Given, \[ \frac{1}{y^2}=\log_e(f(x)) \] Therefore, \[ f(x)=e-4\sin\sqrt{x} \]

Step 6: Final conclusion.
Hence, \[ \boxed{e-4\sin\sqrt{x}} \]