Question

find the solution of the given initial value problem. $$ t y^{\prime}+(t+1) y=t, \quad y(\ln 2)=1 $$

Short answer

Question: Solve the initial value problem for the given first-order linear ordinary differential equation: $$ t y^{\prime}+(t+1) y=t, \quad y(\ln 2)=1 $$ Answer: The solution to the given initial value problem is: $$ y(t) = \frac{1}{t} + \frac{2(\ln 2 - 1)}{e^{t- \ln 2} t} $$

Step-by-step solution

Identify the type of equation

The given equation is: $$ t y^{\prime}+(t+1) y=t, \quad y(\ln 2)=1 $$ This is a first-order linear ordinary differential equation. We can rewrite the equation to make it more explicit: $$ y^{\prime} + \frac{t+1}{t} y = \frac{1}{t} $$

Solving the equation

The next step is to find the integrating factor. The integrating factor is given by: $$ \mu(t) = e^{\int P(t) dt} = e^{\int \frac{t+1}{t} dt} $$ Find the integral: $$ \int \frac{t+1}{t} dt = \int \frac{t}{t} dt + \int \frac{1}{t} dt = \int 1 dt + \int \frac{1}{t} dt = t+\ln t + C $$ Considering only the essential part of the integral without the constant, we get the integrating factor: $$ \mu(t) = e^{t+\ln t} = e^t t $$ Now, we multiply the original equation by the integrating factor: $$ e^t t (y^{\prime} + \frac{t+1}{t} y) = e^t t \cdot \frac{1}{t} $$ Which simplifies to: $$ e^t t y^{\prime}+ (t+1)e^t y = e^t $$ This is now an exact equation in the form \(\frac{d}{dt}(\mu(t) y(t)) = e^t\). Integrate both sides with respect to \(t\): $$ \int d(\mu(t) y(t)) = \int e^t dt $$ Resulting in: $$ e^t t y(t) = e^t+C $$ Now, divide both sides by \(e^t t\): $$ y(t) = \frac{1}{t} + \frac{C}{e^t t} $$

Apply the initial condition

Substitute the initial condition \(y(\ln 2) = 1\) into the solution to find the constant \(C\): $$ 1 = \frac{1}{\ln 2} + \frac{C}{e^{\ln 2} \ln 2} $$ Solve for \(C\): $$ C = (1 - \frac{1}{\ln 2})(e^{\ln 2} \ln 2) = e^{\ln 2} \ln 2 - e^{\ln 2} $$ So, by simplifying, the constant C becomes: $$ C = e^{\ln 2}(\ln 2 - 1) $$ Finally, substitute \(C\) back into the expression for \(y(t)\): $$ y(t) = \frac{1}{t} + \frac{e^{\ln 2}(\ln 2 - 1)}{e^t t} $$ Thus, the solution to the given initial value problem is: $$ y(t) = \frac{1}{t} + \frac{2(\ln 2 - 1)}{e^{t- \ln 2} t} $$

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of differential equation that involves the derivative of a function. Specifically, it is linear because it can be expressed in the form:

• \(y' + P(t)y = Q(t)\)

where \(y'\) denotes the derivative of \(y\) with respect to \(t\), \(P(t)\) and \(Q(t)\) are functions of \(t\).
The linear property makes the equation easier to solve using standard techniques. In our specific problem, we identify it in the form:\[t y^{\prime}+(t+1) y=t\]
To solve it, we often rearrange it to better fit the standard form, as seen by converting it to:

\( y^{\prime} + \frac{t+1}{t} y = \frac{1}{t} \).By rewriting, we transform the equation into a more familiar form, further simplifying the solving process.

Integrating Factor

The integrating factor is a clever method to solve first-order linear differential equations. It is a function by which we multiply the entire differential equation to make one side an exact derivative. This allows us to integrate easily.
In the differential equation form \( y' + P(t)y = Q(t) \), the integrating factor \( \mu(t) \) is computed as follows:

• \( \mu(t) = e^{\int P(t) \, dt} \)

For our problem, we have \( P(t) = \frac{t+1}{t} \), leading to:

• \( \mu(t) = e^{\int \frac{t+1}{t} dt} = e^{t+\ln t} = e^t t \)

This step is crucial as it transforms an equation, making the solution clear through direct integration of both sides.
Multiplying the entire equation by \( \mu(t) \) yields an equation resembling the derivative of a product, facilitating an integration step that simplifies finding the solution.

Exact Equation

An exact equation arises when a differential equation can be written in the form \( \frac{d}{dt}(M(t)y(t)) = N(t) \). Essentially, it means that the differential equation describes the derivative of some function \( M(t)y(t) \).
When we apply the integrating factor to our differential equation, we transform it into this exact form, which is easier to solve. Here's how it works in our problem:

• Start with: \( e^t t y^{\prime} + (t+1)e^t y = e^t \)
• This can be expressed as: \( \frac{d}{dt}(e^t t y) = e^t \)

Now, to solve it, we integrate both sides with respect to \(t\).
By doing this, we exploit the property of exactness to directly integrate and find the general form of the solution.
A substantial benefit is that this process circumvents the need for complex manipulations, leveraging the structure of differential equations.

Initial Condition

The initial condition is a specific point's value that allows us to determine the constant of integration present in a general solution of a differential equation. It helps provide a unique solution that fits specific criteria.
The initial condition in our problem is given by \( y(\ln 2) = 1 \). This tells us that when \( t = \ln 2 \), \( y \) should be \( 1 \).
This condition molds our general solution into the specific one we need.

• Substitute the initial condition into the general solution, \( y(t) = \frac{1}{t} + \frac{C}{e^t t} \), to find \( C \).
• Apply values from the initial condition \( y(\ln 2) = 1 \), yields: \( 1 = \frac{1}{\ln 2} + \frac{C}{e^{\ln 2} \ln 2} \).
• Solve for \( C \) which ensures that this specific point satisfies our equation.

Initial conditions are crucial for generating a solution that applies not just broadly, but precisely to the problem's context. They single out the exact scenario we are interested in from an otherwise infinite set of possibilities.