Toronto Math Forum
MAT2442018S => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii 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 $y(1)=1$.

my answer to question 1

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 $xy$.
$${N_xM_y\over M}={8x2x\over 2xy}={3 \over y}$$
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 $\mu_1(y)=exp^{\int{p(y)}dy}=e^{3lny}=y^3$ $\\$
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 $$x^2y^4+y^4 e^y=e+1$$

(4) is correct but then integrated with an error

(4) is correct but then integrated with an error
Typos. Corrections done.