Question

A particular solution of the differential equation \(\frac{dy}{dx}=xy^2\) with y(0)=1 is

• A. \(y=\frac{2-x^2}{2}\)
• B. \(y=\frac{2}{2-x^2}\)
• C. \(y=\frac{2}{x^2}-2\)
• D. \(y=\frac{x^2-2}{2}\)
• E. \(y=\frac{2}{x^2-2}\)

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{dy}{dx} = xy^2 \] with the initial condition \( y(0) = 1 \).
Step 1: Separation of Variables We first separate the variables: \[ \frac{dy}{y^2} = x \, dx \]
Step 2: Integrate Both Sides Now, integrate both sides: \[ \int \frac{1}{y^2} \, dy = \int x \, dx \] The integral of \( \frac{1}{y^2} \) is \( -\frac{1}{y} \) and the integral of \( x \) is \( \frac{x^2}{2} \), so: \[ -\frac{1}{y} = \frac{x^2}{2} + C \]
Step 3: Solve for \( y \) Now, solve for \( y \): \[ \frac{1}{y} = -\frac{x^2}{2} - C \] \[ y = \frac{1}{-\frac{x^2}{2} - C} \] To simplify the expression, rewrite it as: \[ y = \frac{2}{2 - x^2 - 2C} \]
Step 4: Use the Initial Condition We are given that \( y(0) = 1 \). Substituting \( x = 0 \) and \( y = 1 \) into the equation: \[ 1 = \frac{2}{2 - 2C} \] Solving for \( C \): \[ 2 - 2C = 2 \quad \Rightarrow \quad C = 0 \]

The correct option is (B) : \(y=\frac{2}{2-x^2}\)

Answer 2

We are given the differential equation: \[\frac{dy}{dx} = xy^2\] with the initial condition \(y(0) = 1\).

This is a separable differential equation. We can rewrite it as: \[\frac{dy}{y^2} = x \, dx\]

Now, we integrate both sides: \[\int \frac{dy}{y^2} = \int x \, dx\] \[\int y^{-2} \, dy = \int x \, dx\] \[\frac{y^{-1}}{-1} = \frac{x^2}{2} + C\] \[-\frac{1}{y} = \frac{x^2}{2} + C\]

Now, we use the initial condition \(y(0) = 1\) to find the value of \(C\): \[-\frac{1}{1} = \frac{0^2}{2} + C\] \[-1 = 0 + C\] \[C = -1\]

Substitute \(C = -1\) back into the equation: \[-\frac{1}{y} = \frac{x^2}{2} - 1\] \[\frac{1}{y} = 1 - \frac{x^2}{2} = \frac{2 - x^2}{2}\]

Now, we solve for \(y\): \[y = \frac{2}{2 - x^2}\]

Therefore, the particular solution of the differential equation is: \[y = \frac{2}{2 - x^2}\]