Question

What is the general solution of the equation \(y^{\prime}(t)=-4 y+6 ?\)

Short answer

Question: Solve the differential equation \(y^{\prime}(t)=-4 y+6\). Answer: The general solution of the given differential equation is \(y(t) = Ae^{-4t} + \frac{3}{2}\), where A is a constant.

Step-by-step solution

Identify the type of differential equation

The given equation is a first-order linear nonhomogeneous differential equation with constant coefficients. It can be rewritten in the standard form as \(y^{\prime}(t) + 4y(t) = 6\).

Solve the homogeneous equation

We first need to find the general solution of the homogeneous equation, which is \(y^{\prime}(t) + 4y(t) = 0\). To solve this equation, we can try to find the function \(y(t)\) that satisfies it, in the form \(y(t) = Ae^{kt}\). Differentiating \(y(t)\), we get \(y^{\prime}(t) = Ake^{kt}\). Now we substitute our function and its derivative into the homogeneous equation: \(Ake^{kt} + 4Ae^{kt} = 0\). Factoring out \(Ae^{kt}\), we find \(Ae^{kt}(k + 4) = 0\). Since \(Ae^{kt}\) can never be zero, our equation reduces to \(k + 4 = 0\). Solving for k, we get \(k = -4\). Thus, the general solution of the homogeneous equation is \(y_h(t) = Ae^{-4t}\), where A is a constant.

Find a particular solution

Now we need to find a particular solution to the nonhomogeneous equation \(y^{\prime}(t) + 4y(t) = 6\). We can guess a particular solution in the form of \(y_p(t) = B\), where B is a constant. Differentiating \(y_p(t)\), we get \(y_p^{\prime}(t) = 0\). Substituting \(y_p(t)\) and \(y_p^{\prime}(t)\) into the nonhomogeneous equation, we find: \(0 + 4B = 6\). Solving for B, we find \(B = \frac{3}{2}\). Thus, our particular solution to the nonhomogeneous equation is \(y_p(t) = \frac{3}{2}\).

Write the general solution

The general solution to the nonhomogeneous equation is the sum of the homogeneous solution and the particular solution, i.e. \(y(t) = y_h(t) + y_p(t)\). Substituting the values, we find: \(y(t) = Ae^{-4t} + \frac{3}{2}\). This is our final answer for the general solution of the given differential equation.

Key concepts

first-order linear differential equation

A first-order linear differential equation is an equation of the form \(y'(t) + p(t)y(t) = q(t)\). This means it involves the first derivative of a function \(y\) with respect to \(t\), a variable.

These equations are called 'linear' because they exhibit the property that the dependence of the solution on any initial value is linear. Think of it like a straight road connecting inputs to outputs.

In our example, the equation \(y^{\prime}(t) + 4y(t) = 6\) fits this form, with \(p(t) = 4\) and \(q(t) = 6\). Here, \(t\) is usually considered the independent variable, and \(y(t)\) as the dependent variable, symbolized by its connection with derivatives.

homogeneous solution

A homogeneous solution refers to solving the part of the differential equation that equals zero. In simpler terms, it's like finding a solution where there's no external input or influence on the system.

The equation for the homogeneous part from our exercise is \(y^{\prime}(t) + 4y(t) = 0\). Solving this involves finding a function \(y(t)\) that satisfies it on its own.

Typically, the solution may take the form \(y_h(t) = Ae^{kt}\), where \(A\) is a constant and \(k\) is determined by the equation. This shows that for each point in time, \(y\) changes at a rate proportional to its current value. Solving the equation gives \(y_h(t) = Ae^{-4t}\), exhibiting an exponential decay thanks to the negative exponent.

particular solution

Finding a particular solution involves determining a specific solution that satisfies the nonhomogeneous differential equation. It helps in accounting for the effects of additional inputs or external forces.

In our example, the equation being solved is \(y^{\prime}(t) + 4y(t) = 6\). This time, we aim to find a particular function that fits this exact form.

One common approach is to guess a simple form such as \(y_p(t) = B\), where \(B\) is a constant. After plugging it into the equation, you'd find that \(B = \frac{3}{2}\) satisfies the equation. Thus, the particular solution is \(y_p(t) = \frac{3}{2}\). This complements the homogeneous solution and aids in constructing the general solution.

constant coefficients

When dealing with differential equations, the term 'constant coefficients' refers to the multiplier of \(y(t)\) or its derivative in the equation being a constant number, rather than a function of \(t\).

In the equation \(y^{\prime}(t) + 4y(t) = 6\), the coefficient of \(y(t)\) is 4, which does not change with \(t\).

This feature makes analysis easier since the coefficients remain stable no matter the value of the independent variable \(t\). It simplifies solving as the method does not need to adapt to variations in coefficients over time.

• Constant coefficients allow use of straightforward methods like the characteristic equation for homogeneous equations.
• They simplify the guesswork for particular solutions since the homogeneity and constant inputs remain predictable.

By understanding this, tackling such differential equations becomes more approachable and less of a puzzle.