Question

The solution of \( (x+\log y)dy+ydx=0 \) when \( y(0)=1 \) is

• A. \( y(x-1+\log y)+1=0 \)
• B. \( xy+y\log y+1=0 \)
• C. \( xy-y\log y-y=1 \)
• D. \( y(x+1+\log y)-1=0 \)

Hint:

When a differential equation contains \( dx \) and \( dy \), try to identify whether it becomes a linear differential equation by treating \( x \) as a function of \( y \).

Answer 1

The Correct Option is A

Step 1: Write the given differential equation.
The given differential equation is:
\[ (x+\log y)dy+ydx=0. \]

Step 2: Rearrange the equation.
We can write it as:
\[ ydx=-(x+\log y)dy. \]
Dividing by \( y\,dy \), we get:
\[ \frac{dx}{dy}=-\frac{x+\log y}{y}. \]

Step 3: Convert into linear differential equation form.
Now rearrange the equation:
\[ \frac{dx}{dy}+\frac{x}{y}=-\frac{\log y}{y}. \]
This is a linear differential equation in \( x \) with respect to \( y \).

Step 4: Find the integrating factor.
Here,
\[ P(y)=\frac{1}{y}. \]
So, the integrating factor is:
\[ I.F.=e^{\int \frac{1}{y}dy}=e^{\log y}=y. \]

Step 5: Multiply the equation by integrating factor.
Multiplying the equation by \( y \), we get:
\[ y\frac{dx}{dy}+x=-\log y. \]
The left side is the derivative of \( xy \) with respect to \( y \):
\[ \frac{d}{dy}(xy)=-\log y. \]

Step 6: Integrate both sides.
Integrating, we get:
\[ xy=-\int \log y\,dy. \]
Using the formula \( \int \log y\,dy=y\log y-y \), we get:
\[ xy=-(y\log y-y)+C. \]
Therefore,
\[ xy=-y\log y+y+C. \]

Step 7: Apply the initial condition \( y(0)=1 \).
Since \( y(0)=1 \), we have \( x=0 \) and \( y=1 \).
Substitute these values:
\[ 0=-1\log 1+1+C. \]
Since \( \log 1=0 \),
\[ 0=1+C. \]
So,
\[ C=-1. \]
Thus,
\[ xy=-y\log y+y-1. \]
Bringing all terms to one side:
\[ xy+y\log y-y+1=0. \]
Taking \( y \) common:
\[ y(x-1+\log y)+1=0. \]
Final Answer:
Therefore, the required solution is:
\[ \boxed{y(x-1+\log y)+1=0}. \]