Question

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

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$

Short answer

The solution for the given initial value is $y=\left[\frac{2}{3}x-\frac{2}{9}\right]+\frac{5}{9}{e}^{3x}$ .

Step-by-step solution

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

Consider the following initial value problem:

$\frac{dy}{dx}=2x-3y,y\left(0\right)=\frac{1}{3}$

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,

$$\begin{gathered} \frac{dy}{dx}=2x-3y \\ \frac{dy}{dx}+3y=2x.........\left(1\right) \end{gathered}$$

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

$$\begin{gathered} \frac{dy}{dx}+P\left(x\right)y=Q\left(x\right) \\ P\left(x\right)=3 \end{gathered}$$

The integrating factor is,

$$\begin{gathered} e^{\int p\left(x\right)dx}=e^{\int 3dx} \\ =e^{\int 3dx} \\ =e^{3x} \end{gathered}$$

Thus, the integrating factor is,

$e^{\int p\left(x\right)dx}=e^{3x}$

Determine the general solution for the given differential equation

Multiply the differential equation (1) with integrating factor $e^{\int p\left(x\right)dx}=e^{3x}$

$$\begin{gathered} \Rightarrow e^{3x}\left(\frac{dy}{dx}+3y\right)=2x{e}^{3x} \\ \Rightarrow \frac{d}{dx}\left(y{e}^{3x}\right)dx=2x{e}^{3x}dx \end{gathered}$$

Now, integrate on both sides and solve for x.

$$\begin{gathered} \int \frac{d}{dx}\left(y{e}^{3x}\right)dx=2\int x{e}^{3x}dx \\ \int d\left(y{e}^{3x}\right)=2\int x{e}^{3x}dx \\ \Rightarrow y{e}^{3x}=2\left[\frac{1}{3}x{e}^{3x}-\frac{1}{3}\int e^{3x}dx\right]+C \\ \Rightarrow y{e}^{3x}=\left[\frac{2}{3}x{e}^{3x}-\frac{2}{3}{e}^{3x}dx\right]+C \\ \Rightarrow y=\left[\frac{2}{3}x-\frac{2}{9}\right]+C{e}^{3x} \end{gathered}$$

Therefore, the general solution of the differential equation is $y=\left[\frac{2}{3}x-\frac{2}{9}\right]+C{e}^{3x}$.

$$\begin{gathered} y=\left[\frac{2}{3}x-\frac{2}{9}\right]+C{e}^{3x} \\ \Rightarrow \frac{1}{3}=\left[\frac{2}{3}\left(0\right)-\frac{2}{9}\right]+C{e}^{3\left(0\right)} \\ \Rightarrow \frac{1}{3}=-\frac{2}{9}+C \\ \Rightarrow C=\frac{1}{3}+\frac{2}{9} \\ \Rightarrow C=\frac{5}{9} \end{gathered}$$

Substitute $C=\frac{5}{9}$in the general solution $y=\left[\frac{2}{3}x-\frac{2}{9}\right]+\frac{5}{9}{e}^{3x}$, obtain the solution for initial value problem as,$$\begin{gathered} y=\left[\frac{2}{3}x-\frac{2}{9}\right]+C{e}^{3x} \\ y=\left[\frac{2}{3}x-\frac{2}{9}\right]+\frac{5}{9}{e}^{3x} \end{gathered}$$

Therefore, the solution to the initial value problem (1) is $y=\left[\frac{2}{3}x-\frac{2}{9}\right]+\frac{5}{9}e$

The general solution $y=\left[\frac{2}{3}x-\frac{2}{9}\right]+\frac{5}{9}{e}^{3x}$is a polynomial, so it defined for all real numbers Hence, the largest interval in which the solution defined is $-\infty <y<\infty$.