Question

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

$\mathbf{y}\frac{\mathbf{dx}}{\mathbf{dy}}\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{5}$

Short answer

The solution for the given initial value is $x=2y^2-\frac{49}{5}y$ .

Step-by-step solution

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

Consider the following initial value problem:

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,

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

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

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

The integrating factor is,

$$\begin{gathered} e^{\int \left(y\right)dy}=e^{\int \frac{1}{y}dy} \\ =e^{-\int \frac{1}{y}dy} \\ =e^{-\ln y} \\ =e^{\ln y^{-1}} \\ =y^{-1}\text{Since}{e}^{e^{\ln x}}=x. \\ =\frac{1}{y} \end{gathered}$$

Thus, the integrating factor is,

$e^{\int p\left(y\right)dy}=\frac{1}{y}$

Determine the general solution for the given differential equation

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

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

Now, integrate on both sides and solve for x .

$$\begin{gathered} \int \frac{d}{dy}\left(\frac{x}{y}\right)dy=\int 2dy \\ \Rightarrow \frac{x}{y}=2y+C \\ \Rightarrow x=y(2y+C) \\ \Rightarrow x=2y^2+Cy \end{gathered}$$

Therefore, the general solution of the differential equation is $x=2y^2+Cy$

Use the initial condition y(1) = 5 . to find the value of C.

Substitute x = 1 and Y = 5 in $x=2y^2+Cy$.

$$\begin{gathered} \Rightarrow 1=2(5{)}^2+C\left(5\right) \\ \Rightarrow 1=2\left(25\right)+5C \\ \Rightarrow 1=50+5C \\ \Rightarrow 5C=-49 \\ \Rightarrow C=-\frac{49}{5} \end{gathered}$$

Substitute $C=-\frac{49}{5}$

Substitute C=5 in the general solution $x=2y^2+Cy$, obtain the solution for initial value problem as,

$$\begin{gathered} x=2y^2+Cy \\ x=2y^2-\frac{49}{5}y \end{gathered}$$

Therefore, the solution to the initial value problem (1) is$x=2y^2-\frac{49}{5}y$

The general solution $x=2y^2-\frac{49}{5}y$ is a polynomial, so it defined for all real numbers Hence, the largest interval in which the solution defined is $-\infty <y<\infty$.