Question

First find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the indicated condition. $$ \frac{d y}{d x}=-y^{2} x\left(x^{2}+2\right)^{4} ; y=1 \text { at } x=0 $$

Short answer

The particular solution is \( y = \frac{10}{(x^2 + 2)^5} \).

Step-by-step solution

Separate Variables

The given differential equation is \( \frac{dy}{dx} = -y^2 x (x^2 + 2)^4 \). To solve this equation by separation of variables, we can rewrite it as \( \frac{dy}{y^2} = -x(x^2 + 2)^4 \, dx \). This allows us to integrate both sides separately.

Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \).\[ \int \frac{dy}{y^2} = - \int x(x^2 + 2)^4 \, dx \]The integral of \( \frac{1}{y^2} \) with respect to \( y \) is \( -\frac{1}{y} \), and we will need to apply integration techniques to compute the right integral.

Evaluate the Integrals

The integration of the left side gives:\[ -\frac{1}{y} + C_1 \]For the right side, use substitution (e.g., let \( u = x^2 + 2 \), then \( du = 2x \, dx \)) to solve the integral for the right-hand side. The solution is computed as:\[ -\frac{1}{2} \int u^4 \, du = -\frac{1}{2} \frac{u^5}{5} + C_2 = -\frac{(x^2 + 2)^5}{10} + C_2 \]

Substitute Back and Combine Constants

Combine the results of integration:\[ -\frac{1}{y} = -\frac{(x^2 + 2)^5}{10} + C \]Here, we combined the constants \( C_1 \) and \( C_2 \) into a single constant \( C \) since any constant of integration can be absorbed into another.

Solve for General Solution

Rearrange the expression to solve for \( y \):\[ \frac{1}{y} = \frac{(x^2 + 2)^5}{10} + C \]Thus, the general solution is:\[ y = \frac{10}{(x^2 + 2)^5 + 10C} \]

Apply Initial Condition for Particular Solution

Use the initial condition \( y = 1 \) when \( x = 0 \) to find the particular value of \( C \).Substitute \( y = 1 \) and \( x = 0 \) into the equation:\[ 1 = \frac{10}{(0^2 + 2)^5 + 10C} \]Solve for \( C \):\[ 10 + 10C = 10 \Rightarrow C = 0 \]

Particular Solution

Substitute \( C = 0 \) back into the general solution:\[ y = \frac{10}{(x^2 + 2)^5} \]This is the particular solution that satisfies the initial condition.

Key concepts

Separation of Variables

The method of separation of variables is a fundamental technique for solving differential equations. By manipulating the equation, we attempt to isolate terms involving different variables on opposite sides of the equation. For the given differential equation \( \frac{dy}{dx} = -y^2 x (x^2 + 2)^4 \), separation begins by rewriting it to separate \( y \)-terms from \( x \)-terms: \( \frac{dy}{y^2} = -x(x^2 + 2)^4 \, dx \). This separation allows us to treat one side as a function of \( y \) and the other as a function of \( x \).

• This essential step transforms the differential equation into a pair of integrals.
• It forms the foundation for solving the equation through integration techniques.
• Understanding where and how to perform the separation is crucial for successfully applying this method.

In summary, separation of variables is a pivotal strategy, crucial for solving non-linear differential equations by dividing the problem into simpler, integrable parts.

Integration Techniques

Once the variables are separated, we utilize integration techniques to solve each side of the separated equation. The equation becomes two distinct integrals: \( \int \frac{dy}{y^2} = - \int x(x^2 + 2)^4 \, dx \).

• The left-side integral, \( \int \frac{1}{y^2} \, dy \), simplifies easily to \( -\frac{1}{y} \).
• The right-side integral, more complex, benefits from substitution. Letting \( u = x^2 + 2 \) and \( du = 2x \, dx \), the integral becomes \( -\frac{1}{2} \int u^4 \, du \), which solves to \( -\frac{1}{2} \frac{u^5}{5} \).

Good use of integration techniques like substitution or partial fractions can simplify solving differential equations. These techniques are critical when dealing with more complicated integrals as they help identify a path to a simpler solution.

Initial Conditions

Initial conditions play an essential role in finding particular solutions to differential equations. Rather than just any solution, initial conditions specify unique values that the solution must satisfy, such as \( y = 1 \) when \( x = 0 \) in our problem.

• The initial condition is used to determine the constant of integration that arises during the solution process.
• Once we have the general solution, \( y = \frac{10}{(x^2 + 2)^5 + 10C} \), we substitute the initial condition values in.

By doing this, we find \( C = 0 \), which allows us to define the particular solution. Initial conditions are indispensable in narrowing the infinite possibilities of general solutions to the one which meets specific criteria set by the problem.

Particular Solution

The particular solution of a differential equation is determined after applying initial conditions to the general solution. It represents a precise instance of infinitely many solutions that satisfy the initial condition.

• For our equation, combining the initial condition \( y = 1 \) at \( x = 0 \) resulted in the particular solution \( y = \frac{10}{(x^2 + 2)^5} \).
• The particular solution shows how specific initial values influence the behavior of a solution.
• It is particularly useful in modeling scenarios where initial values at a starting point are available and crucial.

In essence, the particular solution translates the theoretical solution into a practically applicable result, reflecting the real-world system described by the differential equation.