### Author Topic: P1-Day  (Read 3015 times)

#### Victor Ivrii

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##### P1-Day
« on: February 15, 2018, 05:05:37 PM »
Find integrating factor and then a general solution of ODE
\begin{equation*}
2xy +(4x^2+4e^y+ye^y)y'=0 \ .
\end{equation*}

Also, find a solution satisfying $y(1)=1$.

#### Guanlin Song

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##### Re: P1-Day
« Reply #1 on: February 15, 2018, 07:22:24 PM »

#### Meng Wu

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##### Re: P1-Day
« Reply #2 on: February 15, 2018, 11:52:19 PM »
Let $$M(x,y)=2xy, N(x,y)=4x^2+4e^y+ye^y$$
Check exactness:
$${\partial{M} \over \partial{y}}=2x, {\partial{N} \over \partial{x}}=8x$$
Since, ${\partial{M} \over \partial{y}} \neq {\partial{N} \over \partial{x}}$, we need to find an integrating factor $\mu(x,y)$. $\\$
Now check if $\mu(x,y)$ if depends on only $x$, or depends on only $y$, or depends on only $xy$.
$${N_x-M_y\over M}={8x-2x\over 2xy}={3 \over y}$$
Thus the integrating factor $\mu(x,y)$ is depends on only $y$ which is $\mu(y)={ 3\over y}$. $\\$
Hence $${d\mu\over dy}={3\over y}\mu \implies \mu'-{3\over y}\mu=0$$
Integrating factor $\mu_1(y)=exp^{\int{p(y)}dy}=e^{3ln|y|}=y^3$ $\\$
Now we multiply $\mu_1(y)=y^3$ to both sides of the equation:
\begin{align}y^3\cdot 2xy+y^3\cdot(4x^2+4e^y+ye^y)&=0 \\ 2xy^4+(4x^2y^3+4y^3e^y+y^4e^y)y' &=0\end{align}
Now the differential equation becomes Exact:$\\$
So there is a function $\psi(x,y)$ such that:
\begin{align}\psi_x(x,y)&=2xy^4 \\ \psi_y(x,y)&=4x^2 y^3+4y^3e^y+y^4e^y\end{align}
We integrate $\psi_x(x,y)$ with respect to $x$:
$$\psi(x,y)=x^2y^4+h(y)$$
Hence,\begin{align}\psi_y(x,y)&=4x^2y^3+h'(y)\\&=4x^2y^3+4y^3e^y+y^4e^y\end{align} and
$$h'(y)=4y^3e^y+y^4e^y \implies h(y)=y^4 e^y$$
Therefore, $$\psi(x,y)=x^2y^4+y^4 e^y$$ and the general solution of this ODE is $$x^2y^4+y^4 e^y=c$$
Now let $x=1$ and $y=1$:
$$(1)^2 (1)^4+(1)^4 e^1=c$$
Therefore, the solution of the IVP is $$x^2y^4+y^4 e^y=e+1$$
« Last Edit: February 21, 2018, 08:37:30 AM by Meng Wu »

#### Victor Ivrii

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##### Re: P1-Day
« Reply #3 on: February 21, 2018, 08:17:00 AM »
(4) is correct but then integrated with an error

#### Meng Wu

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##### Re: P1-Day
« Reply #4 on: February 21, 2018, 08:31:10 AM »
(4) is correct but then integrated with an error

Typos. Corrections done.