WBJEE · Maths · Differential Equations
General solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0\) is \((C\) is an arbitrary constant)
- A \(\frac{x}{a}=\tan \frac{y}{a}+C\)
- B \(\tan x y=C\)
- C \(\tan (x+y)=C\)
- D \(\tan \frac{y+C}{a}=\frac{x+y}{a}\)
Answer & Solution
Correct Answer
(D) \(\tan \frac{y+C}{a}=\frac{x+y}{a}\)
Step-by-step Solution
Detailed explanation
\((x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0\) Let \(x+y=t\) \(1+\frac{d y}{d x}=\frac{d t}{d x}\) \(\Rightarrow \quad \frac{d y}{d x}=\frac{d t}{d x}-1\) Therefore, \(t^{2}\left(\frac{d t}{d x}-1\right)=a^{2}\) \(\Rightarrow \quad t^{2} \frac{d t}{d x}-t^{2}=a^{2}\)…
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