Question

Find a particular solution satisfying the given conditions. \(x y^{\prime}-y=x^{2}, \quad y=6\) when \(x=2\)

Short answer

The particular solution is \( y = x^2 + x \).

Step-by-step solution

- Rewrite the differential equation

Start by rewriting the given differential equation in a more recognizable form. The given equation is: \[ x y' - y = x^2 \] Add \( y \) to both sides to isolate the derivative term: \[ x y' = y + x^2 \] Now, divide both sides by \( x \) to make the equation in the form of a standard first-order linear differential equation: \[ y' - \frac{y}{x} = x \]

- Identify the integrating factor

The standard form of a first-order linear differential equation is: \[ y' + p(x)y = q(x) \] In our case, \(p(x) = -\frac{1}{x}\) and \(q(x) = x\). The integrating factor is found using the formula: \[ \mu(x) = e^{\int p(x) dx} \] Calculate \( \int -\frac{1}{x} dx \): \[ \int -\frac{1}{x} dx = -\ln|x| \] So the integrating factor becomes: \[ \mu(x) = e^{-\ln|x|} = |x|^{-1} = \frac{1}{x} \]

- Multiply the differential equation by the integrating factor

Multiply every term in the differential equation by the integrating factor \( \frac{1}{x} \): \[ \frac{1}{x} y' - \frac{y}{x^2} = 1 \] Notice this simplifies to: \[ \left( \frac{y}{x} \right)' = 1 \]

- Integrate both sides

Integrate both sides with respect to \( x \): \[ \int \left( \frac{y}{x} \right)' dx = \int 1 dx \] This results in: \[ \frac{y}{x} = x + C \] where \( C \) is the constant of integration.

- Solve for the particular solution

We need to solve for \( y \). Multiply both sides by \( x \) to get: \[ y = x^2 + Cx \] Use the given initial condition, \( y = 6 \) when \( x = 2 \), to find \( C \): \[ 6 = 2^2 + 2C \] Simplify to solve for \( C \): \[ 6 = 4 + 2C \] \[ 2 = 2C \] \[ C = 1 \]

- Write the final particular solution

Substitute \( C = 1 \) back into the general solution: \[ y = x^2 + x \] Therefore, the particular solution satisfying the given conditions is: \( y = x^2 + x \)

Key concepts

Integrating Factor

When solving first-order linear differential equations, the integrating factor is a crucial tool. It allows us to turn a differential equation into a simpler form that is easier to solve. The general form of a first-order linear differential equation is:\[ y' + p(x)y = q(x) \] The integrating factor, denoted as \( \mu(x) \), is derived from the function \( p(x) \). It is calculated by taking the exponential of the integral of \( p(x) \):\[ \mu(x) = e^{\int p(x) dx} \] In our example, first rewrite the given equation \( x y' - y = x^2 \) in standard form to identify \( p(x) \) and \( q(x) \):\[ y' - \frac{y}{x} = x \] Here, \( p(x) = -\frac{1}{x} \). Calculating the integral:\[ \int -\frac{1}{x} dx = -\ln|x| \]This gives us the integrating factor:\[ \mu(x) = e^{-\ln|x|} = |x|^{-1} = \frac{1}{x} \] Multiplying the original differential equation by this integrating factor helps simplify and solve it efficiently.

Differential Equation Solution

After finding the integrating factor, the goal is to simplify the differential equation. For our example, multiplying by \( \frac{1}{x} \) results in:\[ \frac{1}{x} y' - \frac{y}{x^2} = 1 \] This can be rewritten as:\[ \( \frac{y}{x} \)' = 1 \] Now, we integrate both sides:\[ \int \( \frac{y}{x} \)' dx = \int 1 dx \] The left side of the integral simplifies to the function inside the derivative:\[ \frac{y}{x} = x + C \] To find \( y \), multiply both sides by \( x \):\[ y = x^2 + Cx \] This represents the general solution to the differential equation. Using initial conditions helps to find the particular solution specific to our problem.

Initial Conditions

Initial conditions provide specific values for the variables at a given point, which help pinpoint a unique solution to a differential equation. In our problem, we use the initial condition \( y = 6 \) when \( x = 2 \). Plugging these into the general solution:\[ y = x^2 + Cx \] we get:\[ 6 = 2^2 + 2C \] Simplifying to find \( C \):\[ 6 = 4 + 2C \] \[ 2 = 2C \] \[ C = 1 \] Substitute \( C \) back into the general solution:\[ y = x^2 + x \] This yields the particular solution which satisfies both the differential equation and the initial condition. Initial conditions are key to defining a specific solution and ensure the equation models real-world problems accurately.