Question

$$ (3 x+8)\left(y^{2}+4\right) d x-4 y\left(x^{2}+5 x+6\right) d y=0, \quad y(1)=2 $$

Short answer

The solution of the given differential equation with the initial condition \(y(1) = 2\) is: \[2y^2 (x^2 + 5x + 6) = (3x+8)y^3 + 4(3x+8)y\]

Step-by-step solution

Rewrite the equation in the form of separable variables

The given differential equation can be rewritten in the following form: \[ \frac{dy}{dx} = \frac{(3x + 8)(y^2 + 4)}{4y(x^2 + 5x + 6)} \]

Separate the variables

Now we can separate the variables by multiplying both sides with the denominators containing x and y: \[ 4y(x^2 + 5x + 6) dy = (3x + 8)(y^2 + 4) dx \]

Integrate both sides

We integrate both sides of the equation with respect to their respective variables: \[ \int 4y(x^2 + 5x + 6) dy = \int (3x + 8)(y^2 + 4) dx \] Performing the integration, we get: \[ 2y^2 (x^2 + 5x + 6) = (3x+8)y^3 + 4(3x+8)y + C \]

Apply the initial condition

We now apply the initial condition, \(y(1) = 2\), to the equation to find the constant C: \[ 2(2)^2 (1^2 + 5(1) + 6) = (3(1) + 8)(2)^3 + 4(3(1) + 8)(2) + C \] Solving for C, we get: \[ C = 0 \]

Write down the solution

Finally, we can rewrite the equation as the solution of the given differential equation: \[ 2y^2 (x^2 + 5x + 6) = (3x+8)y^3 + 4(3x+8)y \] This is the solution of the given first-order differential equation with the specified initial condition.

Key concepts

Separable Differential Equations

Separable differential equations are a special type of differential equations where you can separate the variables on each side of the equation. This makes them relatively easier to solve compared to other types of differential equations. To recognize a separable differential equation, check if you can write it in the form:\[N(y)\frac{dy}{dx} = M(x)\]

• Here, \( N(y) \) is a function of \( y \) only.
• \( M(x) \) is a function of \( x \) only.

In the given exercise, you initially have an equation that looks complicated, but the first step involves rewriting it to separate \( x \) and \( y \). This manipulation is crucial as it transforms the differential equation into integrable forms. Once separated, each side of the equation can be integrated with respect to its own variable, leading towards the solution.

Initial Value Problems

An initial value problem (IVP) involves solving a differential equation together with a specific value, called the initial condition, which gives you additional information about the solution. This initial condition is usually given in the form:\[y(x_0) = y_0\]

• \( x_0 \) is the particular value of \( x \) where the condition is specified.
• \( y_0 \) is the value of \( y \) at that specific \( x_0 \).

The purpose of the initial condition is to allow us to find the constant of integration. In this exercise, the initial condition provided is \( y(1) = 2 \). By substituting this initial value into the equation after integration, you can solve for the constant \( C \). This step is critical; without applying the initial condition, the solution would remain arbitrary with an unknown constant.

Integration of Differential Equations

Integration is a fundamental step in solving separable differential equations. Once you have separated the variables, integration allows you to find the actual solutions. The goal is to perform the integration on both sides:\[\int N(y)\, dy = \int M(x)\, dx\]

• For the \( y \) side, integrate with respect to \( y \).
• For the \( x \) side, integrate with respect to \( x \).

In this exercise, after rearranging the equation, you find integrals on both sides. The integration process might require different techniques such as substitution, partial fractions, or simply applying standard integral formulas. The result of integration gives you a family of solutions, typically involving an arbitrary constant \( C \). Later, using the initial condition as mentioned, you find the particular value of \( C \), refining the solution to satisfy the original differential equation and its initial requirement.