Search any question & find its solution
Question:
Answered & Verified by Expert
The general solution of the differential equation $\left(\frac{d y}{d x}\right)+y \cdot g^{\prime}(x)=g(x) \cdot g^{i}(x)$, where $g(x)$ is a given function of $x$ is
Options:
Solution:
1682 Upvotes
Verified Answer
The correct answer is:
$g(x)+\log (1+y-g(x))=c$
Given, differential equation is
$$
\frac{d y}{d x}+y \cdot g^{\prime}(x)=g(x) \cdot g^{\prime}(x)
$$
This is in the form of linear differential equation.
So, $\quad \mathrm{IF}=e^{\int g^{\prime}(x) d x}=e^{g(x)}$
Required solution is
$$
y \cdot e^{g(x)}=\int g(x) \cdot g^{\prime}(x) \cdot e^{g(x)} d x+c^{\prime}
$$
Put $g(x)=t \Rightarrow g^{\prime}(x) d x=d t$
$$
\therefore y \cdot e^{g(x)}=\int t e^{t} d t+c^{\prime}
$$
$$
\begin{aligned}
&\Rightarrow y \cdot e^{g(x)}=t e^{t}-e^{t}+c^{\prime} \\
&\Rightarrow y \cdot e^{g(x)}=(g(x)-1) e^{g(x)}+c^{\prime} \\
&\text { or }(y-g(x)+1) e^{g(x)}=c^{\prime}
\end{aligned}
$$
Taking log on both sides, we get
$$
g(x)+\log (1+y-g(x))=c
$$
Where, $c=\log c^{\prime}$
$$
\frac{d y}{d x}+y \cdot g^{\prime}(x)=g(x) \cdot g^{\prime}(x)
$$
This is in the form of linear differential equation.
So, $\quad \mathrm{IF}=e^{\int g^{\prime}(x) d x}=e^{g(x)}$
Required solution is
$$
y \cdot e^{g(x)}=\int g(x) \cdot g^{\prime}(x) \cdot e^{g(x)} d x+c^{\prime}
$$
Put $g(x)=t \Rightarrow g^{\prime}(x) d x=d t$
$$
\therefore y \cdot e^{g(x)}=\int t e^{t} d t+c^{\prime}
$$
$$
\begin{aligned}
&\Rightarrow y \cdot e^{g(x)}=t e^{t}-e^{t}+c^{\prime} \\
&\Rightarrow y \cdot e^{g(x)}=(g(x)-1) e^{g(x)}+c^{\prime} \\
&\text { or }(y-g(x)+1) e^{g(x)}=c^{\prime}
\end{aligned}
$$
Taking log on both sides, we get
$$
g(x)+\log (1+y-g(x))=c
$$
Where, $c=\log c^{\prime}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.