Question

Solve the following differential equations by power series and also by an elementary method. Verify that the series solution is the power series expansion of your other solution. $$ x y^{\prime}=x y+y $$

Short answer

y = C_1 x e^x (series) matches with y = C_1 x e^x (exponential).

Step-by-step solution

Rewrite the Differential Equation

Rewrite the given differential equation in standard form. The given equation is: \[ x y^{\text{'}} = x y + y \]Divide both sides by x to get: \[ y^{\text{'}} = y + \frac{y}{x} \]

Separate Variables

Rewrite the equation to separate variables, putting all terms involving y on one side and all terms involving x on the other. \[ y^{\text{'}} = y + \frac{y}{x} = \left(1 + \frac{1}{x}\right) y \]Thus, \[ \frac{dy}{y} = \left(1 + \frac{1}{x}\right) dx \]

Integrate Both Sides

Integrate both sides of the separated equation.\[ \int \frac{1}{y} dy = \int \left(1 + \frac{1}{x}\right) dx \]This gives:\[ \ln|y| = x + \ln|x| + C \]

Solve for y

Exponentiate both sides to solve for y.\[ e^{\ln|y|} = e^{x + \ln|x| + C} \]This simplifies to:\[ y = e^{x} |x| e^{C} \]Letting \(e^C = C_1\), the solution is:\[ y(x) = C_1 x e^{x} \]

Power Series Solution Form

Assume a power series solution of the form\[ y = \sum_{n=0}^{\infty} a_n x^n \]Then, the derivative is\[ y^{\text{'}} = \sum_{n=1}^{\infty} n a_n x^{n-1} \]

Substitute Series into Differential Equation

Substituting these into the differential equation, we get:\[ x \sum_{n=1}^{\infty} n a_n x^{n-1} = x \sum_{n=0}^{\infty} a_n x^n + \sum_{n=0}^{\infty} a_n x^n \]Simplify and combine terms:\[ \sum_{n=1}^{\infty} n a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+1} + \sum_{n=0}^{\infty} a_n x^n \]

Align Indexes and Compare Coefficients

Align indexes and compare coefficients of like powers of x. Set the coefficients of each power of x to zero:For \(x^0\): \[ a_0 = 0 \]For \( x^{n+1}\): \[ (n+1) a_{n+1} = a_n \]Therefore, \[ a_{n+1} = \frac{a_n}{n+1} \]and due to recursive nature, general term:\[ a_n = \frac{a_1}{n!} \]

Form Solution

Given the general term, the power series is:\[ y = \sum_{n=1}^{\infty} \frac{a_1}{n!} x^n = a_1 \sum_{n=1}^{\infty} \frac{x^n}{n!} \]This simplifies using known series for the exponential function:\[ y = a_1 (e^x - 1) \]Since an arbitrary constant can be redefined, let \(a_1 = C_1\).\[ y = C_1 (e^x - 1) \]

Verify Power Series with Exponential Solution

Verify that the obtained power series solution matches the original solution.Original exponential solution:\[ y = C_1 x e^x \]Power series solution:\[ y = C_1 x e^x \]Both methods give a consistent solution.

Key concepts

power series

To understand how power series are used in solving differential equations, we begin with the concept itself. A power series is an infinite series of the form: \[ y = \sum_{n=0}^{\infty} a_n x^n \] where \( a_n \) are coefficients and \( x^n \) are powers of \( x \).
This allows us to express a function as an infinite sum of terms in powers of \( x \).
In the context of solving differential equations, we assume that the solution can be expressed as a power series.
We then find the coefficients \( a_n \) step by step, using given initial conditions and other constraints.
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For example, if you start with \( y = \sum_{n=0}^{\infty} a_n x^n \), its derivative is \( y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \).
Substituting these into the differential equation, we align indexes, compare coefficients, and solve for the coefficients \( a_n \).
This approach helps in finding a series representation for the solution.

separation of variables

Separation of variables is a key technique to solve differential equations.
The idea is to rewrite the equation in such a way that all terms involving one variable appear on one side of the equation, and all terms involving the other variable appear on the other side.
Consider the given equation: \[ x y^{\'} = x y + y \] By dividing both sides by \( x \), we get: \( y^{\'} = y + \frac{y}{x} \).Next, rewriting this to separate variables, we have \( \frac{dy}{y} = (1 + \frac{1}{x})dx \).
Once the variables are separated, we integrate both sides to find the solution.
This process involves two main steps: separating the variables and then integrating both sides.
By integrating both sides, we often end up with a natural logarithm or an exponential expression, which we can solve further to find the function \( y \).

exponential function

The exponential function \( e^x \) plays a crucial role in solving differential equations.
It's a key function, often arising when solving equations involving growth or decay processes.
In our differential equation solution, after integrating both sides, we obtained \ln|y| = x + \ln|x| + C.\This simplification involves exponentiating both sides:
\[ e^{\ln|y|} = e^{x + \ln|x| + C} \]
The laws of exponents allow us to write this as:

\( y = e^x |x| e^C \)
Letting \( e^C = C_1 \), the solution simplifies to \( y = C_1 x e^x \).
The exponential function thus provides a way to simplify and solve the differential equation, leading us to the final solution.

integration

Integration is the process of finding the integral of a function.
It's used extensively to solve differential equations.
In our exercise, we separated the variables to integrate both sides of the equation:
\( \int \frac{1}{y} dy = \int (1 + \frac{1}{x}) dx \)
This step-by-step integration allows us to find \ln|y| = x + \ln|x| + C.\Once we perform the integration, we solve for the arbitrary constant \( C \) using initial conditions.
This leads us closer to the final solution of the differential equation.
In this exercise, after integrating both sides, we determined that \( y = C_1 x e^x \).
Integrating properly helps us transition from a differential equation to finding a function that satisfies the given conditions.