Question

If \(y=A \mathrm{e}^{2 x}\) is the general solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}=2 y\), find the particular solution satisfying \(y(0)=3\). What is the particular solution satisfying \(\frac{\mathrm{d} y}{\mathrm{~d} x}=2\) when \(x=0 ?\)

Short answer

Question: Find the particular solutions for the given differential equation, \(\frac{dy}{dx}=2y\), with the given conditions: 1) \(y(0)=3\), and 2) \(\frac{dy}{dx}=2\) when \(x=0\). Answer: The particular solutions are: 1) \(y=3e^{2x}\) and 2) \(y=e^{2x}\).

Step-by-step solution

Substitute x=0 and y=3 into the general solution

We have \(y=Ae^{2x}\), and we know that \(y(0)=3\). So, we substitute \(x=0\) and \(y=3\) in the given formula: \(3=Ae^{2\cdot0}\) which simplifies to \(3=Ae^0\). Now, we know that \(e^0=1\). Therefore, we get: \(3=A\cdot1\) or \(A=3\). #Step 2: Obtain the particular solution for y(0)=3#

Replace A in the general solution

Now that we have found A, we can substitute it in the general solution: \(y=3e^{2x}\). So, the particular solution satisfying \(y(0)=3\) is: \(y=3e^{2x}\). Now let's find the second particular solution. #Step 3: Find the constant A for the second condition when dy/dx=2 and x=0#

Calculate dy/dx and substitute x=0 and dy/dx=2

First, we calculate the derivative of the general solution: \(\frac{dy}{dx}= \frac{d}{dx} (Ae^{2x}) = A\cdot2e^{2x}.\) Next, we substitute \(x=0\) and \(\frac{dy}{dx}=2\): \(2=A\cdot(2e^{2\cdot0})=2Ae^0\). Now, we know that \(e^0=1\). Therefore, we get: \(2=2A\cdot1\) or \(A=1\). #Step 4: Obtain the particular solution for dy/dx=2 when x=0 #

Replace A in the general solution

Now, we have found A for the second case. We can substitute it in the general solution: \(y=e^{2x}\). So, the particular solution satisfying \(\frac{dy}{dx}=2\) when \(x=0\) is: \(y=e^{2x}\).

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. They are broadly used in disciplines such as physics, engineering, economics, and biology.

Differential equations can be simple or complex and can be classified into several types. The most common types are ordinary differential equations (ODEs) and partial differential equations (PDEs). In the context of our problem, we're dealing with an ODE because it involves functions of only one variable and its derivatives.

Exponential Functions

Exponential functions are fundamental in mathematics and appear often in real-world applications, such as in the growth of populations, in finance to compute compound interest, and in physics to describe decay processes. An exponential function can be written in the form \( y = A \times e^{kx} \), where \( A \) and \( k \) are constants, and \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

These functions are unique as their rate of growth is proportional to their value, which is why they naturally describe processes where growth accelerates over time or decays exponentially. The problem provided showcases an exponential function where the general solution is in the given form, with a constant multiple A, which we are aiming to determine.

Initial Value Problems

An initial value problem (IVP) is a type of differential equation along with an initial condition. The initial condition specifies the value of the unknown function at a certain point, which is necessary to find a unique solution to the differential equation — the particular solution. Without this initial condition, the best we can find is a general solution that includes an arbitrary constant.

In the exercise at hand, we are given the initial conditions \( y(0)=3 \) and \( \frac{dy}{dx}=2 \) when \( x=0 \). These initial conditions allow us to find the specific values of A that satisfy the given conditions, thus providing us with the particular solutions for those specific scenarios.

Derivatives

Derivatives represent the rate at which a function is changing at any given point and are foundational in calculus. For a function \( y=f(x) \), the derivative is typically denoted \( \frac{dy}{dx} \) or \( f'(x) \). It measures how sensitive a function is to small changes in the input value, which in physical terms can correspond to velocity, acceleration, or slope, depending on the context.

In our problem, we differentiate an exponential function which results in a similar exponential expression, revealing the exponential function's unique property that its derivative is proportional to the function itself. The proportional constant is determined by the exponent in the given function, which, once identified, allows for solving initial value problems.