Question

A special class of first-order linear equations have the form $a(t)y^{\prime }(t)+a^{\prime }(t)y(t)=f(t),$ where $a$ and $f$ are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t)y^{\prime }(t)+a^{\prime }(t)y(t)=\frac{d}{dt}(a(t)y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to $t.$ Use this idea to solve the following initial value problems. $$(t^2+1)y^{\prime }(t)+2ty=3t^2,y(2)=8$$

Short answer

Answer: The specific solution of the given first-order linear equation with the initial condition is $y(t)=\frac{t^3+5t^2+5}{t^2+1}$.

Step-by-step solution

Identify the product derivative

We notice that the left side of the given equation can be written as the derivative of a product: $$(t^2+1)y^{\prime }(t)+2ty(t)=\frac{d}{dt}(a(t)y(t))$$ where $a(t)=t^2+1$. To confirm this, we take the derivative of $a(t)y(t)$, which gives us: $$\frac{d}{dt}(a(t)y(t))=(t^2+1)y^{\prime }(t)+2ty(t)$$

Integrate both sides

Now, we integrate both sides with respect to $t$ to solve the equation: $$\int \frac{d}{dt}(a(t)y(t))dt=\int f(t)dt$$ $$\int (t^2+1)y^{\prime }(t)+2ty(t)dt=\int 3t^{2}dt$$ On integrating, we have: $$a(t)y(t)=\int 3t^{2}dt+C$$ where $C$ is the constant of integration.

Substitute $a(t)$ and solve for $y(t)$

Now, let's substitute $a(t)=t^2+1$ back into our equation: $$(t^2+1)y(t)=\int 3t^{2}dt+C$$ Integrating $3t^2$ with respect to $t$ yields: $$(t^2+1)y(t)=t^3+C$$ Now let's solve for $y(t)$ by dividing both sides by $t^2+1$: $$y(t)=\frac{t^3}{t^2+1}+\frac{C}{t^2+1}$$

Use the initial condition to find the constant $C$

We are given the initial condition $y(2)=8$. Plug in this value into the equation we derived above: $$8=\frac{2^3}{2^2+1}+\frac{C}{2^2+1}$$ Solving for $C$, we obtain: $$C=5(t^2+1)$$

Write the final solution

Now, plug in the value of $C$ into the equation for $y(t)$: $$y(t)=\frac{t^3}{t^2+1}+\frac{5(t^2+1)}{t^2+1}$$ We can simplify the equation as: $$y(t)=\frac{t^3+5(t^2+1)}{t^2+1}$$ So, the final solution is: $$y(t)=\frac{t^3+5t^2+5}{t^2+1}$$

Key concepts

Initial Value Problem

An initial value problem in the context of differential equations involves finding a specific solution to a differential equation given an initial condition. In our exercise, we deal with the differential equation

• $(t^2+1)y^{\prime }(t)+2ty(t)=3t^2$

The initial condition provided is: $y(2)=8$. This condition helps to uniquely determine the constant of integration, $C$, once the general solution is found.

By using the initial condition, we can substitute back into our final function for $y(t)$ and solve for $C$. This substitution ensures the solution satisfies not only the differential equation but also the specific value at the initial point. Ensuring consistency with the initial condition is crucial in solving initial value problems.

Integration

Integration is used in solving differential equations by finding antiderivatives. In the given exercise, once we recognize that the left-hand side of our equation can be expressed as a derivative of a product, we are tasked with integrating both sides of the equation.

• Since the equation is $\frac{d}{dt}(a(t)y(t))=f(t)$, integrating both sides with respect to $t$ gives us a way to simplify and solve for $a(t)y(t)$.
• On the right side, integrating $f(t)=3t^2$ with respect to $t$ results in $t^3+C$.

By integrating, we move from the rate of change (represented by derivatives) to obtaining a function that represents accumulated quantities or values over an interval, essential in finding our specific function $y(t)$. Integration is a fundamental tool in calculus that provides the ability to transition from knowing the rate at which something changes to understanding the total amount or value of what changes.

Derivative of a Product

The derivative of a product rule is a formula used to find the derivative of two multiplicative functions. In our exercise, the derivative of a product is key because it helps us to reframe our differential equation.

• The left-hand side $(t^2+1)y^{\prime }(t)+2ty(t)$ is the result of applying the derivative of a product rule on $a(t)y(t)$ leading to: $\frac{d}{dt}((t^2+1)y(t))$.
• This recognition allows us to simplify the problem into an integration task.

Understanding the product rule means comprehending that when differentiating a product $uv$, the result is $u^{\prime }v+u{v}^{\prime }$. In the exercise, $(t^2+1)$ represents $u$ and $y(t)$ represents $v$, allowing us to structure the given differential equation in a form that is easier to integrate and solve.