Question

The solution of (x+y+1)dy = dx is
[a] $x+y+2=C{{e}^{y}}$
[b] $x+y+4=C\log y$
[c] $\log \left( x+y+1 \right)=Cy$
[d] $\log \left( x+y+2 \right)=C+{{y}^{2}}$

Answer 1

Hint: Divide both sides of the equation by dy and hence prove that the given equation is equivalent to the differential equation $\dfrac{dx}{dy}-x=y+1$. Use the fact that the solution of the differential equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ is given by $yIF=\int{Q\left( x \right)IFdx}$, where the integrating factor IF is given by $IF={{e}^{\int{P\left( x \right)dx}}}$
Hence find the integrating factor of the above equation and hence solve the differential equation.

Complete step-by-step answer:
We have
$\left( x+y+1 \right)dy=dx$
Dividing both sides by dy, we get
$\dfrac{dx}{dy}=x+y+1$
Subtracting x from sides, we get
$\dfrac{dx}{dy}-x=y+1$, which is of the form $\dfrac{dx}{dx}+P\left( y \right)x=Q\left( y \right)$ and hence is a linear differential equation with the dependent variable as x and independent variable as y.
We know that the solution of the differential equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ is given by $yIF=\int{Q\left( x \right)IFdx}$, where the integrating factor IF is given by $IF={{e}^{\int{P\left( x \right)dx}}}$
Here P(y) = -1 and Q(y) = y+1
Hence, we have
$IF={{e}^{\int{-1dy}}}={{e}^{-y}}$ and the solution of the equation is given by
$x{{e}^{-y}}=\int{\left( y+1 \right){{e}^{-y}}dy}$
We know that if $\dfrac{df\left( x \right)}{dx}=u\left( x \right)$ and $\int{g\left( x \right)}dx=v\left( x \right)$, then $\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)v\left( x \right)-\int{u\left( x \right)v\left( x \right)dx}$
This is known as integration by parts rule. f(x) is known as the first function and g(x) as the second function.
Now, we know that $\dfrac{d}{dy}\left( y+1 \right)=1$ and $\int{{{e}^{-y}}dy}=-{{e}^{-y}}$
Hence, using integration by parts taking $y+1$ as the first function and ${{e}^{-y}}$ as the second function, we get
$\int{\left( y+1 \right){{e}^{-y}}}=-\left( y+1 \right){{e}^{-y}}+\int{{{e}^{-y}}}=-\left( y+1 \right){{e}^{-y}}-{{e}^{-y}}=-{{e}^{-y}}\left( y+2 \right)$
Hence, we have
$x{{e}^{-y}}=-{{e}^{-y}}\left( y+2 \right)+C$, where C is a constant of integration
Multiplying both sides by ${{e}^{y}}$, we get
$x=-y-2+C{{e}^{y}}$
Hence, we have
$x+y+2=C{{e}^{y}}$, which is the required solution of the differential equation.
Hence option [a] is correct.

Note: Alternative solution
Put x+y+1 =t
Differentiating both sides with respect to y, we get
$\begin{aligned} & \dfrac{dt}{dy}=\dfrac{dx}{dy}+1 \\\ & \Rightarrow \dfrac{dx}{dy}=\dfrac{dt}{dy}-1 \\\ \end{aligned}$
Hence, we have
$\dfrac{dt}{dy}-1=t$
Adding 1 on both sides, we get
$\dfrac{dt}{dy}=t+1$
Hence, we have
$\dfrac{dt}{t+1}=dy$
Integrating both sides, we get
$\ln \left( t+1 \right)=y+a$, where a is a constant
Hence, we have
$\begin{aligned} & t+1={{e}^{y}}{{e}^{a}} \\\ & \Rightarrow t+1=C{{e}^{y}} \\\ \end{aligned}$
where $C={{e}^{a}}$
Reverting to original variables, we get
$\begin{aligned} & x+y+1+1=C{{e}^{y}} \\\ & \Rightarrow x+y+2=C{{e}^{y}} \\\ \end{aligned}$
Hence option [a] is correct.