Question

A particular solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}=(x+9 y)^2\), when \(x=0, y=\frac{1}{27}\) is

• A. \(3x+27y=\mathrm{tan}[3(x+\frac{\pi}{12})]\)
• B. \(3x+27y=\mathrm{tan}(x+\frac{\pi}{4})\)
• C. \(3x+27y=\mathrm{tan}(x+\frac{\pi}{12})\)
• D. \(3x+27y=\mathrm{tan}[3(x+\frac{\pi}{4})]\)

Hint:

Substitute u = x + 9y to convert the equation into one with separable variables.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Let \(u=x+9y\). Then \(\dfrac{du}{dx}=1+9\dfrac{dy}{dx}=1+9u^2\).

Step 2: Separate and Integrate
\(\displaystyle\int\dfrac{du}{1+9u^2}=\int dx\Rightarrow\dfrac{1}{3}\tan^{-1}(3u)=x+c\)

Step 3: General Solution
\(3u=\tan[3(x+c)]\), that is \(3x+27y=\tan[3(x+c)]\).

Step 4: Apply the Condition
At \(x=0,\ y=\dfrac{1}{27}\): \(1=\tan(3c)\Rightarrow3c=\dfrac{\pi}{4}\Rightarrow c=\dfrac{\pi}{12}\). So the solution is \(3x+27y=\tan[3(x+\tfrac{\pi}{12})]\).

Final Answer:
The particular solution is 3x + 27y = tan[3(x + pi/12)]. \[ \boxed{\text{(A) }3x+27y=\tan\left[3\left(x+\dfrac{\pi}{12}\right)\right]} \]