Question

Solve the given initial value problem. Give the largest interval lover which the general solution is defined.

$\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{8}$

Short answer

The solution for the given initial value is $y=2x+1+5x^{-1}$.

Step-by-step solution

The given equation in the standard form and determine the integrated factor

Consider the following initial value problem:

$x\frac{dy}{dx}+y=4x+1,y\left(1\right)=8$

The objective is to solve the following initial value problem (IVP) and give the largest interval over which the solution is defined. Rewrite the given differential equation as,

$\frac{dy}{dx}+\frac{y}{x}=\frac{4x+1}{x}$ …….. (1)

Compare the given differential equation with the linear equation of the form

$$\begin{gathered} \frac{dy}{dx}+P\left(x\right)y=Q\left(x\right) \\ P\left(x\right)=\frac{1}{x},Q\left(x\right)=\frac{4x+1}{x} \end{gathered}$$

The integrating factor is,

$$\begin{gathered} e^{\int p\left(x\right)dx}=e^{\int \frac{1}{x}dx} \\ =e^{\int \frac{1}{x}dx} \\ =e^{\ln x} \\ =e^{\ln x^1} \\ =x\text{Since}{e}^{e^{\ln x}}=x \end{gathered}$$

Thus, the integrating factor is, $e^{\int p\left(x\right)dx}=x$

Determine the general solution for the given differential equation

Multiply the differential equation (1) with integrating factor $e^{\int p\left(x\right)dx}=x$

$$\begin{gathered} \Rightarrow x\left(\frac{dy}{dx}+\frac{1}{xy}\right)=x\left(\frac{4x+1}{x}\right) \\ \Rightarrow x\frac{dy}{dx}+\frac{1}{y}=4x+1 \end{gathered}$$

Now, integrate on both sides and solve for .

Therefore, the general solution of the differential equation is x.

$$\begin{gathered} \int d\left(yx\right)=\int (4x+1)dx \\ \Rightarrow yx=2x^2+x+C \\ \Rightarrow y=\frac{2x^2+x+C}{x} \end{gathered}$$

Use the initial condition y(1) =8 . to find the value of C.

Substitute x = 1 and y = 8 in $y=\frac{2x^2+x+C}{x}$.

$$\begin{gathered} y=\frac{2x^2+x+C}{x} \\ \Rightarrow 8=\frac{2{\left(1\right)}^2+1+C}{1} \\ \Rightarrow 8-3=C \\ \Rightarrow C=5 \end{gathered}$$

$$\begin{gathered} y=\frac{2x^2+x+C}{x} \\ y=\frac{2x^2+x+5}{x} \\ y=2x+1+5x^{-1} \end{gathered}$$

Therefore, the solution to the initial value problem (1) is $y=2x+1+5x^{-1}$

The general solution is a polynomial, so it defined for all real numbers Hence, the largest interval in which the solution defined is $-\infty <y<\infty$.