Question

Find the general solution of the given differential equation. Give the largest intervall over which the general solution is defined. Determine whether there are any transient terms in the general solution.

${\mathbf{yd}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{(}\mathbf{x}\mathbf{+}{\mathbf{y}}^{\mathbf{k}}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

The general solution of given differential equation is $x=2y^6+C{y}^4$and

there are no any transient terms in the general solution.

Step-by-step solution

Rewriting the given equation in the form

Consider the differential equation,

$y{d}^2-4\left(x+y^k\right)dy=0$

Rewrite the differential equation,

$$\begin{gathered} ydx=4\left(x-y^6\right)dy \\ \frac{dx}{dy}=\frac{4\left(x+y^6\right)}{y} \\ \frac{dx}{dy}=\frac{4x}{y}+\frac{4y^6}{y} \\ \frac{dx}{dy}-\frac{4}{y}x=4y^5.........\left(1\right) \end{gathered}$$

Compare this equation with the standard form of linear differential equation in the variable x

$$\begin{gathered} \frac{dx}{dy}=P\left(y\right)x=Q\left(y\right) \\ P\left(y\right)=-\frac{4}{y}\text{and}P\left(y\right)=4y^5 \end{gathered}$$

So,

Note than P and Q are continuous on $(-\infty ,0)~(0,\infty )$.

Determine the integrating factor

Now find the integrating factor_{$e^{\int P\left(y\right)dy}$}

$$\begin{gathered} e^{\int P\left(y\right)dy}=e^{\int \frac{-4}{y}dy} \\ \Rightarrow e^{-4\ln y} \\ \Rightarrow e^{\ln y^{-4}} \\ \Rightarrow y^{-4} \end{gathered}$$

Determine the general solution of given differential equation

Multiply the equation (1) with an integrating factor y^-4.

$$\begin{gathered} y^{-4}\frac{dx}{dy}-\frac{4}{y^5}x=4y \\ \frac{d}{dy}\left(\frac{x}{y^4}\right)-4y \\ \left\{\frac{d}{dy}\left(\frac{1}{y^4}\right)-y^{-4}\frac{dx}{dy}-\frac{4}{y^5}x\right\} \end{gathered}$$

Integrate both sides with respect an y yields to

$$\begin{gathered} \int \frac{d}{dy}\left(\frac{x}{y^4}\right)=\int 4y+C \\ \frac{x}{y^4}=\frac{4}{2}{y}^2+C \\ \frac{x}{y^4}=2y^2+C \\ x=y^4\left(2y^2+C\right) \\ x=2y^6+C{y}^4 \end{gathered}$$

Therefore, the general solution of given differential equation is $x=2y^6+C{y}^4$

Here,$x>0 fory\in (-\infty ,0)\cup (0,\infty )$

Thus, the general solution is defined in the interval

$I=(0,\infty )$

The objective is to determine whether there are any transient terms in the general solution

Since transient term of general solution of differential equation is term that not approaches to zero as X approaches to infinity.

Therefore, solution is not Transient.