Question

Find the solution of \[\left( {x + y + 1} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = 1\].
A. \[y{\rm{ }} = {\rm{ }}(x{\rm{ }} + {\rm{ }}2){\rm{ }} + {\rm{ }}c{e^x}\]
B. \[y{\rm{ }} = {\rm{ }} - {\rm{ }}(x{\rm{ }} + {\rm{ }}2){\rm{ }} + {\rm{ }}c{e^x}\]
C. \[x{\rm{ }} = {\rm{ }} - {\rm{ }}(y{\rm{ }} + {\rm{ }}2){\rm{ }} + {\rm{ }}c{e^y}\]
D. \[x{\rm{ }} = {\rm{ }}{(y{\rm{ }} + {\rm{ }}2)^2} + {\rm{ }}c{e^y}\]

Answer 1

Hint: substitute the given expression as \[x{\rm{ }} + {\rm{ }}y{\rm{ }} + 1 = {\rm{ }}u\] and then, \[du{\rm{ }} = {\rm{ }}dx{\rm{ }} + {\rm{ }}dy\]. Then, after putting the values, integrate both the sides and solve the integral further.

Formula used:
1. \[\int {\dfrac{1}{x}dx} = \log \left| x \right| + C\]
2. \[\int {dx} = x + C\]

Complete step by step solution:
We have the given equation is: \[\left( {x + y + 1} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = 1\]
Now,
\[\left( {x{\rm{ }} + {\rm{ }}y{\rm{ }} + {\rm{ }}1} \right){\rm{ }}dy{\rm{ }} = {\rm{ }}dx\] ………………………..(1)
We will assume it as equation (1).
Substituting \[x{\rm{ }} + {\rm{ }}y{\rm{ }} + 1 = {\rm{ }}u\], we have \[du{\rm{ }} = {\rm{ }}dx{\rm{ }} + {\rm{ }}dy\] and now the given equation reduces to
\[u\left( {du - dx} \right){\rm{ }} = {\rm{ }}dx\;\]
\[ \Rightarrow \dfrac{{{\rm{udu}}}}{{{\rm{u + 1}}}}\] = \[dx\]
\[ \Rightarrow \dfrac{{{\rm{(u+1-1)du}}}}{{{\rm{u + 1}}}}\] = \[dx\]
Now, on integrating on both the sides, we get:
\[ \Rightarrow \int {\left( {{\rm{1 - }}\dfrac{{\rm{1}}}{{{\rm{u + 1}}}}} \right)} {\rm{du = }}\int {{\rm{dx}}} \]
\[ \Rightarrow {\rm{u - log(u + 1) = x + k}}\]
Putting the value of \[x{\rm{ }} + {\rm{ }}y{\rm{ }} + 1 = {\rm{ }}u\], we get:
\[ \Rightarrow {\rm{log(x + y + 2) = y + C}}\]
Taking antilog on both sides, we have:
\[ \Rightarrow {\rm{x + y + 2 = C}}{{\rm{e}}^y}\]
The correct answer is option C.

Note-A differential equation is one that has a function and its derivatives. It can be referred to as either an ordinary differential equation (ODE) or a partial differential equation, depending on whether partial derivatives are present or not (PDE).