Question Statement

Solve the differential equation:

$(1-x)dy+y^{-1}dx=0;y(0)=2$

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Background and Explanation

This is a separable differential equation, where we can rearrange the equation to isolate all $y$ terms on one side with $dy$ and all $x$ terms on the other side with $dx$. Once separated, we integrate both sides and use the initial condition to determine the constant of integration.

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Solution

Step 1: Separate the variables

Starting with the given equation: $(1-x)dy+\frac{1}{y}dx=0$

Rearrange to isolate the differential terms: $(1-x)dy=-\frac{1}{y}dx$

Multiply both sides by $y$ and divide by $(1-x)$ to separate variables: $ydy=\frac{-1}{1-x}dx$

Step 2: Integrate both sides

Integrate the left side with respect to $y$ and the right side with respect to $x$:

$\int ydy=\int \frac{-1}{1-x}dx$

Evaluating the integrals:

• Left side: $\int ydy=\frac{y^2}{2}$
• Right side: $\int \frac{-1}{1-x}dx=\ln (1-x)$ (since the derivative of $\ln (1-x)$ is $\frac{-1}{1-x}$)

Thus: $\frac{y^2}{2}=\ln (1-x)+c$

where $c$ is the constant of integration.

Step 3: Apply the initial condition

Given $y(0)=2$, substitute $x=0$ and $y=2$ into equation (1):

$\frac{(2{)}^2}{2}=\ln (1-0)+c$

Simplify: $\frac{4}{2}=\ln (1)+c$

$2=0+c$

Therefore: $c=2$

Step 4: Write the final solution

Substitute $c=2$ back into equation (1):

$\frac{y^2}{2}=\ln (1-x)+2$

This can also be written as: $y^2=2\ln (1-x)+4$

or $y=\sqrt{2\ln (1-x)+4}$

(Note: We take the positive root since $y(0)=2>0$)

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Key Formulas or Methods Used

• Separation of variables: Rearranging $M(x)N(y)dx+P(x)Q(y)dy=0$ into the form $f(y)dy=g(x)dx$
• Power rule for integration: $\int ydy=\frac{y^2}{2}$
• Logarithmic integration: $\int \frac{-1}{1-x}dx=\ln (1-x)+C$ (valid for $x<1$)
• Initial value condition: Substituting $y(x_0)=y_0$ to solve for the constant of integration

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Summary of Steps

1. Rearrange the equation to separate variables: $ydy=\frac{-1}{1-x}dx$
2. Integrate both sides: $\frac{y^2}{2}=\ln (1-x)+c$
3. Substitute the initial condition $y(0)=2$ to find $c=2$
4. Write the particular solution: $\frac{y^2}{2}=\ln (1-x)+2$