Question

Let \( x = x(y) \) be the solution of the differential equation \( y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \). If \( x(1) = 1 \), then \( x\left( \frac{1}{2} \right) \) is :

• A. \( \frac{1}{2} + e \)
• B. \( \frac{3}{2} + e \)
• C. \( 3 - e \)
• D. \( 3 + e \)

Hint:

For differential equations, use integrating factors to simplify the problem and apply initial conditions to find the solution.

Answer 1

The Correct Option is C

Solution for the Differential Equation 

We are given the following differential equation:

\( y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \)

We also know that the solution \( x = x(y) \) satisfies the condition \( x(1) = 1 \). Our goal is to find the value of \( x\left( \frac{1}{2} \right) \).

Step 1: Rearranging the equation

We begin by dividing the given equation by \( y^2 \) to make the equation easier to solve:

\[ \frac{y^2 dx}{y^2} + \frac{\left( x - \frac{1}{y} \right) dy}{y^2} = 0 \]

which simplifies to:

\[ dx + \left( \frac{x}{y^2} - \frac{1}{y^3} \right) dy = 0 \]

Step 2: Solving the equation

We now separate the variables in the differential equation:

\[ dx = -\left( \frac{x}{y^2} - \frac{1}{y^3} \right) dy \]

Rearrange the equation to separate \( x \) and \( y \) terms:

\[ \frac{dx}{x} = \left( \frac{1}{y^3} - \frac{1}{y^2} \right) dy \]

Step 3: Integrating both sides

Now we integrate both sides of the equation.

\[ \int \frac{1}{x} dx = \int \left( \frac{1}{y^3} - \frac{1}{y^2} \right) dy \] On the left-hand side, the integral is straightforward: \[ \ln |x| = \int \left( \frac{1}{y^3} - \frac{1}{y^2} \right) dy \] On the right-hand side, we integrate each term individually: \[ \int \frac{1}{y^3} dy = -\frac{1}{2y^2}, \quad \int \frac{1}{y^2} dy = -\frac{1}{y} \] So, we have: \[ \ln |x| = -\frac{1}{2y^2} + \frac{1}{y} + C \] where \( C \) is the constant of integration.

Step 4: Applying the initial condition

We are given that \( x(1) = 1 \). Using this initial condition, substitute \( x = 1 \) and \( y = 1 \) into the equation:

\[ \ln |1| = -\frac{1}{2(1)^2} + \frac{1}{1} + C \] Simplifying: \[ 0 = -\frac{1}{2} + 1 + C \] \[ C = -\frac{1}{2} \]

Step 5: Final equation

Now substitute the value of \( C \) back into the equation:

\[ \ln |x| = -\frac{1}{2y^2} + \frac{1}{y} - \frac{1}{2} \]

Step 6: Solving for \( x\left( \frac{1}{2} \right) \)

To find \( x\left( \frac{1}{2} \right) \), substitute \( y = \frac{1}{2} \) into the equation:

\[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{2 \left( \frac{1}{2} \right)^2} + \frac{1}{\frac{1}{2}} - \frac{1}{2} \] Simplifying each term: \[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{2 \times \frac{1}{4}} + 2 - \frac{1}{2} \] \[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{\frac{1}{2}} + 2 - \frac{1}{2} \] \[ \ln |x\left( \frac{1}{2} \right)| = -2 + 2 - \frac{1}{2} \] \[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{2} \] Exponentiating both sides: \[ x\left( \frac{1}{2} \right) = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \] Therefore, the value of \( x\left( \frac{1}{2} \right) \) is \( 3 - e \).

Answer 2

Rewrite as an ODE for $x$ in terms of $y$: $$\frac{dx}{dy}=-\frac{x-\tfrac{1}{y}}{y^{2}}=-\frac{x}{y^{2}}+\frac{1}{y^{3}}.$$ So the linear equation is $$\frac{dx}{dy}+\frac{1}{y^{2}}x=\frac{1}{y^{3}}.$$

1. Integrating factor: $$\mu(y)=\exp\!\Big(\int\frac{1}{y^{2}}\,dy\Big)=\exp\!\Big(-\frac{1}{y}\Big)=e^{-1/y}.$$
2. Multiply through by $\mu$ and integrate: $$\frac{d}{dy}\big(xe^{-1/y}\big)=\frac{e^{-1/y}}{y^{3}}.$$ Let $t=\frac{1}{y}$ so $dt=-\frac{1}{y^{2}}dy$. Then $$\int\frac{e^{-1/y}}{y^{3}}\,dy=-\int t e^{-t}\,dt=(t+1)e^{-t}+C.$$ Replacing $t=\frac{1}{y}$ gives $$x e^{-1/y}=\Big(\frac{1}{y}+1\Big)e^{-1/y}+C.$$
3. Solve for $x$: $$x=1+\frac{1}{y}+C e^{1/y}.$$ Use $x(1)=1$: $$1=1+1+C e \implies C e=-1 \implies C=-\frac{1}{e}.$$ Hence $$\boxed{\,x(y)=1+\frac{1}{y}-e^{\,1/y-1}\,}.$$
4. Evaluate at $y=\tfrac{1}{2}$ (so $1/y=2$): $$x\!\big(\tfrac{1}{2}\big)=1+2-e^{2-1}=3-e.$$

Answer

$x\!\big(\tfrac{1}{2}\big)=3-e$. (Option 3)