# Differential equations in function. Equations (1) : xy′+(1-x)y=1 let z=xy+1

Differential equations in function
Equations (1): $x{y}^{\prime }+\left(1-x\right)y=1$
determine and solve the differential equation (2) whose general solution is the function z .
determine the general solution of (1)
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

soporoseun
Step 1
Compute ${z}^{\prime }=y+x{y}^{\prime }=y+1-\left(1-x\right)y=z$
So the general solution of the above equation (call it eqn (2)) is
$z\left(x\right)=z\left(0\right){e}^{x}={e}^{x}$
Step 2
because $z\left(0\right)=1$. Solving then for y:
$y\left(x\right)=\frac{z-1}{x}=\frac{{e}^{x}-1}{x}$
###### Did you like this example?
Gillian Cooper
Step 1
If $z=xy+1$ then
$\begin{array}{rcl}\frac{dz}{dx}& =& \frac{d}{dx}\left(xy+1\right)\\ & =& y+x\frac{dy}{dx}\end{array}$
Step 2
Now:
$\begin{array}{rcl}x\frac{dy}{dx}+y-xy& =& 1\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\frac{dy}{dx}+y& =& 1+xy\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{dz}{dx}& =& z\end{array}$
Now you solve this equation for z and then change back the variables to x and y at the end.