Question

Find the solution to the initial-value problem.\(\frac{d y}{d x}=y^{3} x e^{x^{2}}, y(0)=1\)

Short answer

The solution to the initial-value problem is \( y = \frac{1}{\sqrt{1 - e^{x^2}}} \).

Step-by-step solution

Understand the Form of the Differential Equation

The given differential equation is \(\frac{dy}{dx} = y^3 x e^{x^2}\). It's separable, meaning it can be manipulated into two separate integrals, one involving only \(y\) and the other involving only \(x\).

Separate Variables

Rearrange the terms to isolate the variables: \(y^{-3} dy = x e^{x^2} dx\). This allows us to integrate both sides with respect to their respective variables.

Integrate Both Sides

Integrate the left side with respect to \(y\) and the right side with respect to \(x\):\[ \int y^{-3} \, dy = \int x e^{x^2} \, dx \]This gives:\[-\frac{1}{2} y^{-2} = \frac{1}{2} e^{x^2} + C \]

Solve for the Constant Using the Initial Condition

Apply the initial condition \(y(0) = 1\) to find \(C\). Substitute \(y = 1\) and \(x = 0\) into the equation:\[-\frac{1}{2}(1)^{-2} = \frac{1}{2} e^{0} + C \]\[-\frac{1}{2} = \frac{1}{2} + C \]Therefore, \(C = -1\).

Solve for \(y\)

Substitute \(C = -1\) back into the integrated form:\[-\frac{1}{2} y^{-2} = \frac{1}{2} e^{x^2} - 1 \]Multiply both sides by \(-2\) to solve for \(y^{-2}\):\[y^{-2} = 1 - e^{x^2}\]Take the reciprocal to solve for \(y\):\[y = \pm \frac{1}{\sqrt{1 - e^{x^2}}}\]Since \(y(0) = 1\), \(y\) is positive, so:\[y = \frac{1}{\sqrt{1 - e^{x^2}}}\]

Verify Solution Meets Initial Conditions

Verify that when \(x = 0\), \(y(0) = 1\):\(y = \frac{1}{\sqrt{1 - e^{0}}} = \frac{1}{\sqrt{1 - 1}} = 1\).This confirms the solution meets the initial condition.

Key concepts

Separable Differential Equations

Separable differential equations are a common type of differential equation that can be easily solved through a simple procedure. These equations allow us to separate the variables involved so that we can integrate them separately. This typically involves manipulating the given equation until each side depends on only one variable.
For example, in our problem, the equation \(\frac{dy}{dx} = y^3 x e^{x^2} \) is separable because we can rearrange it to isolate the \(y\) terms on one side and the \(x\) terms on the other: \( y^{-3} dy = x e^{x^2} dx \). This separation is key, allowing us to handle each variable independently through integration later on.
When dealing with separable differential equations, remember:

• Identify if the equation allows the separation of variables.
• Rearrange the equation to isolate terms involving \(y\) on one side and terms involving \(x\) on the other.

Integration Techniques

Once we have separated the variables, integration techniques come into play. Integration is the process of calculating the integral of a function, which is essentially the opposite of differentiation.
In our example, once we have \( y^{-3} dy = x e^{x^2} dx \), we integrate both sides separately. For the left side, integrating \( y^{-3} \) with respect to \( y \) gives us \( -\frac{1}{2} y^{-2} \). This is achieved through a simple power rule integrated in reverse.
For the right side, the integration of \( x e^{x^2} \) with respect to \( x \) can be a bit more complex and might require techniques such as substitution, especially when dealing with the exponential term \( e^{x^2} \). Often, a well-selected substitution can simplify the process greatly.

• Apply the basic rules of integration on simple functions.
• Consider substitutions for more complex functions.
• Simplify your expressions as much as possible before integrating.

Initial Conditions

Initial conditions are crucial in solving differential equations as they allow us to find the particular solution that fits a given situation. An initial condition gives a specific value of the function at an initial point, which helps in determining the constant of integration.
In our problem, the initial condition provided is \( y(0) = 1 \). After integrating, we ended up with an expression containing the constant \( C \). To find \( C \), substitute the initial values \( x = 0 \) and \( y = 1 \) into the integrated equation. This step is essential because it tailors the general solution to meet the specific physical or theoretical context defined by the problem.
Remember:

• Substitute the given initial condition into the integrated form.
• Solve for the constant of integration \( C \).
• Use the value of \( C \) to express the complete particular solution.
• Verify the initial condition is satisfied by substituting it back into your final solution.

These steps ensure the solution meets the required initial conditions, confirming its validity.