# 2.2.5 de equation with y(1)=1/2

• MHB
• karush
In summary, find the solution of the given initial value problem by rewriting the equation and finding u(x) using integration. Then, multiply both sides of the equation by x^2 and integrate again to solve for y. The solution is given by y=\frac{1}{12}(3x^2-4x+6+\frac{1}{x^2}), and the interval of validity is from x=0 to infinity.
karush
Gold Member
MHB
$\textsf{Find the solution of the given initial value problem. State the interval in which the solution is valid.}$
$$xy^\prime+2y=x^2-x+1, \qquad y(1)=\frac{1}{2}$$
$\textit{rewrite}$
$$y' +\frac{2}{x}y=x-1+\frac{1}{x}$$
$\textit {Find u(x) }$
$$\displaystyle u(x)=\exp\int \frac{2}{x} \, dx =\ln e^{x^2}=x^2$$
$\textit{multiply thru with$u(x)=x^2$}$
$$x^2y' +(x^2)'y=x^2(x^2-x+1)$$
$\textsf{rewrite:}$
$$(x^2 y)'=x^4-x^3+x^2 \, \color{red}{\textit{I think the problem is here}}$$
$\textit{Integrate }$
$$\displaystyle x^2y=\int x^4-x^3+x^2 \, dx =\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}+c$$
$\textit{divide thru by$x^2$}$
$$\displaystyle y=\frac{x^3}{5}-\frac{x^2}{4}+\frac{x}{3}+\frac{c}{x^2}$$
$\textit{book answer(I couldn't get this) }$
$$y=\color{red}{\frac{1}{12}(3x^2-4x+6+\frac{1}{x^2})}$$

Last edited:
You have the ODE: $xy'+2y=x^2-x+1$ you should multiply both sides of the equation by $x$ and not x^2, that's where you got it wrong, your RHS is multiplied by x^2 and not x.

$\textit{but isn't$u(x)=x^2$}$

Last edited:
karush said:
$\textit{but isn't$u(x)=x^2$}$

Yes, but you used it on the original ODE, not the one in standard linear form. :)

Ok here is my final flower arraignment

$\textit{Find the solution of the given initial value problem.}$
$\textit{State the interval in which the solution is valid.}$
\begin{array}{lrll}
\textit{Given:}\\
\textit{Divide thru by x}\\
&\displaystyle y' +\frac{2}{x}y&\displaystyle =x-1+\frac{1}{x}& &&(2)\\
\textit {Find u(x)}\\
&\displaystyle u(x)&\displaystyle =\exp\int \frac{2}{x} \, dx =\ln e^{x^2}=x^2&&&(3)\\
\textit{Multiply thru $x^2$}\\
&x^2y' +(x^2)'y&\displaystyle =x^2\left(x-1+\frac{1}{x}\right)&&&(4)\\
\textit{Rewrite:}\\
&(x^2 y)'&=x^3-x^2+x \, &&&(5)\\
\textit{Integrate}\\
&\displaystyle x^2y&\displaystyle =\int x^3-x^2+x \, dx &&&(6)\\
&&\displaystyle=\frac{x^4}{4}-\frac{x^3}{3}+\frac{x^2}{2}+c\\
\textit{Divide thru by $x^2$}\\
&\displaystyle y&\displaystyle=\frac{x^2}{4}-\frac{x}{3}+\frac{1}{2}+\frac{c}{x^2}&&&(7)\\
\textit{Solve for c}\\
&\displaystyle y(1)&\displaystyle=\frac{1}{4}-\frac{1}{3}+\frac{1}{2}+c=\frac{1}{2}&&&(8)\\
\textit{With $\displaystyle c=\frac{1}{12}$ then}\\
&y&=\color{red}{\displaystyle\frac{1}{12}\left(3x^2-4x+6+\frac{1}{x^2}\right)}&&&(9)
\end{array}

suggestions and insults welcome
actually I don't know what the interval is that valid?

## 1. What is the significance of the initial condition y(1)=1/2 in the equation?

The initial condition y(1)=1/2 specifies the value of the dependent variable y at the initial value of the independent variable x=1. This condition helps to uniquely determine the solution to the differential equation, as it acts as a starting point for the solution.

## 2. How does changing the value of the initial condition affect the solution to the equation?

Changing the value of the initial condition will result in a different solution to the equation. This is because the initial condition acts as a starting point for the solution and altering it will change the path that the solution takes.

## 3. Can you explain the significance of the numbers 2.2.5 in the equation?

The numbers 2.2.5 indicate the order and degree of the differential equation. In this case, the equation is a second-order differential equation with a degree of 5. This information helps to determine the complexity and methods needed to solve the equation.

## 4. What methods can be used to solve this type of differential equation?

There are various methods that can be used to solve a second-order differential equation, such as the method of undetermined coefficients, variation of parameters, and Laplace transform. The specific method used will depend on the form and complexity of the equation.

## 5. What are some real-world applications of this type of differential equation?

This type of differential equation can be used to model a wide range of physical phenomena, such as the motion of a mass on a spring, the growth of a population, or the decay of a radioactive substance. It is also commonly used in engineering, economics, and other fields to analyze and predict behavior in dynamic systems.

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