KCET · Maths · Differential Equations
The solution of the differential equation \(\frac{d y}{d x}=(x+y)^2\) is
- A \(\tan ^{-1}(x+y)=x+C\)
- B \(\tan ^{-1}(x+y)=0\)
- C \(\cot ^{-1}(x+y)=C\)
- D \(\cot ^{-1}(x+y)=x+C\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}(x+y)=x+C\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=(x+y)^2\)
Let \(x+y=t \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x}\)
Eq. (i) can be written as
\[
\begin{gathered}
\frac{d t}{d x}-1=t^2 \\
\Rightarrow \quad \frac{d t}{d x}=t^2+1 \Rightarrow \frac{d t}{t^2+1}=d x
\end{gathered}
\]
\(\begin{aligned} & \Rightarrow \int \frac{d t}{t^2+1}=\int d x \Rightarrow \tan ^{-1} t=x+C \\ & \Rightarrow \quad \tan ^{-1}(x+y)=x+C\end{aligned}\)
Let \(x+y=t \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x}\)
Eq. (i) can be written as
\[
\begin{gathered}
\frac{d t}{d x}-1=t^2 \\
\Rightarrow \quad \frac{d t}{d x}=t^2+1 \Rightarrow \frac{d t}{t^2+1}=d x
\end{gathered}
\]
\(\begin{aligned} & \Rightarrow \int \frac{d t}{t^2+1}=\int d x \Rightarrow \tan ^{-1} t=x+C \\ & \Rightarrow \quad \tan ^{-1}(x+y)=x+C\end{aligned}\)
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