Question

Solve the initial value problem. $$ x^{2} y^{\prime}=2 x^{2}+y^{2}+4 x y, \quad y(1)=1 $$

Short answer

Answer: The solution to the given initial value problem is \(y(x) = -2x + y^2 x + \frac{2}{5}x^4\).

Step-by-step solution

Rewrite the given equation in standard form

First, we need to rewrite the given equation in the form \(y'(x) = F(x, y)\). Divide both sides of the equation by \(x^2\) to obtain: $$ y'(x) = 2 + \frac{y^2}{x^2} + \frac{4y}{x} $$

Write the equation as a separable differential equation

Now, we will rewrite the equation in the form \(M(y)dy = N(x)dx\). To do this, we need to separate the variables \(x\) and \(y\). We will rearrange the equation as follows: $$ \frac{dy}{dx} - \frac{4y}{x} = 2 + \frac{y^2}{x^2} $$ Now, move the \(dx\) to the right side of the equation to get: $$ dy - \frac{4y}{x} dx = (2 + \frac{y^2}{x^2}) dx $$

Integrating both sides

Now, we will integrate both sides of the equation with respect to their respective variables, i.e., integrate \(dy\) and \((2 + \frac{y^2}{x^2}) dx\). Since the left side is not yet easy to integrate, apply integrating factor. The integrating factor \(\mu(x)\) is given by \(\mu(x) = e^{\int P(x) dx}\), where \(P(x) = -\frac{4}{x}\): $$ \mu(x) = e^{\int -\frac{4}{x} dx} = e^{-4\ln|x|} = x^{-4} $$ Multiplying the equation by the integrating factor, we get: $$ x^{-4}(dy - \frac{4y}{x} dx) = x^{-4}(2 + \frac{y^2}{x^2}) dx $$ This simplifies to: $$ (x^{-4}y)' = 2x^{-4} + y^2 x^{-6} $$ Now, integrate both sides with respect to x: $$ \int (x^{-4}y)' dx = \int (2x^{-4} + y^2 x^{-6}) dx $$

Finding the function y(x)

After integrating both sides, we get: $$ x^{-4}y = -2x^{-3} -\frac{1}{5}y^2 x^{-5} + C $$ Now, we need to solve for y(x) by multiplying through by \(x^4\) and rearranging: $$ y = -2x + y^2 x + 5Cx^4 $$

Apply the initial condition

To determine the constant of integration \(C\), we will apply the initial condition \(y(1)=1\): $$ 1 = -2(1) + (1^2)(1) + 5C(1^4) $$ Which simplifies to: $$ C = \frac{2}{5} $$

Write the final solution

Finally, substitute the value of \(C\) back into the equation for y(x) to obtain the solution to the initial value problem: $$ y(x) = -2x + y^2 x + \frac{2}{5}x^4 $$ This is the solution to the given initial value problem.

Key concepts

Differential Equations

Differential equations are fundamental in mathematics. They involve functions and their derivatives. Essentially, they describe how a quantity changes over time or space. In our given problem, we have the differential equation \( x^2 y' = 2x^2 + y^2 + 4xy \). This equation relates the derivative \( y' \) of the function \( y \) with respect to \( x \) to \( y \) and \( x \) themselves. By solving differential equations, we find how the function \( y \) behaves or changes.
- Differential equations can model real-world phenomena, like motion, growth, and decay.
- Solving them involves finding a general or specific function that satisfies the equation for given conditions, such as initial or boundary values.
Understanding differential equations is crucial because they form the base for many engineering and science problems.

Separable Equations

Separable equations are a special type of differential equations where variables can be separated into two sides of the equation. This means rearranging the equation so that each variable appears with its differential on a different side. In our case, the equation \( \frac{dy}{dx} - \frac{4y}{x} = 2 + \frac{y^2}{x^2} \) can be manipulated into a separable form. However, the equation was not completely separable without further techniques.
- Solving separable equations typically involves integrating both sides separately.
- The goal is to achieve expressions where you can integrate with respect to each variable independently.
This separation technique simplifies the process of solving differential equations, as it reduces the complexity to basic calculus integration.

Integrating Factor

The integrating factor is a method used to solve linear differential equations that couldn't be tackled with simple solutions or separation. It's particularly useful for first-order linear equations. In our scenario, the differential equation was not easily separated, requiring us to utilize an integrating factor to facilitate integration.
To find the integrating factor, \( \mu(x) \), we identify \( P(x) \) and compute \( e^{\int P(x) dx} \). Here, we found \( \mu(x) = x^{-4} \). This function, once multiplied to each part of the equation, transforms it into a form where the left-hand side becomes a derivative of a product \( (x^{-4}y)' \).
- Multiplying through by the integrating factor ensures both sides of the equation become integrable.
The integrating factor method is a powerful tool that simplifies solving certain differential equations by reducing reliance on guesswork and transforming the equation into a simple integrable form.

Boundary Value Problems

Boundary value problems (BVP) are similar to initial value problems (IVPs), but instead of providing a condition at a single point, BVPs have conditions specified at two or more points. Typically, these involve differential equations with solutions that need to satisfy conditions at specific boundaries.- In an initial value problem, like the one we solved \( y(1)=1 \), only a single initial point condition is provided.
- In BVPs, you might be given conditions at both the beginning and end-points of the domain.
BVPs are common in scenarios like mechanical engineering, where stress might need to meet specific conditions at both ends of a beam. They require different strategies, such as shooting methods or finite difference methods, to solve as the behavior across a domain needs to be considered.