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-The solution of the differential equation $\frac{d y}{d x}=(x+y)^{2}$ is
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Verified Answer
The correct answer is:
$\tan ^{-1}(x+y)=x+c$
We have,
$$
\frac{d y}{d x}=(x+y)^{2}
$$
Let $\quad x+y=z$
$$
\Rightarrow \quad \frac{d y}{d x}+1=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{d z}{d x}-1
$$
Now, given equation becomes
$$
\frac{d z}{d x}-1=z^{2} \Rightarrow \frac{d z}{d x}=1+z^{2}
$$
$$
d x=\frac{d z}{1+z^{2}}
$$
On integrating both sides, we get
$$
\begin{array}{ll}
& \int d x=\int \frac{d z}{1+z^{2}} \\
\Rightarrow \quad & c+x=\tan ^{-1}(z) \\
\Rightarrow \quad & \tan ^{-1} z=x+c \\
\Rightarrow \quad & \tan ^{-1}(x+y)=x+c
\end{array}
$$
$$
\frac{d y}{d x}=(x+y)^{2}
$$
Let $\quad x+y=z$
$$
\Rightarrow \quad \frac{d y}{d x}+1=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{d z}{d x}-1
$$
Now, given equation becomes
$$
\frac{d z}{d x}-1=z^{2} \Rightarrow \frac{d z}{d x}=1+z^{2}
$$
$$
d x=\frac{d z}{1+z^{2}}
$$
On integrating both sides, we get
$$
\begin{array}{ll}
& \int d x=\int \frac{d z}{1+z^{2}} \\
\Rightarrow \quad & c+x=\tan ^{-1}(z) \\
\Rightarrow \quad & \tan ^{-1} z=x+c \\
\Rightarrow \quad & \tan ^{-1}(x+y)=x+c
\end{array}
$$
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