Answer:
(y^2+1)(x^2+1)=k
Step-by-step explanation:
It looks like it is possibly separatable.
Let's try.
Factor the x out on first term and factor the y out on second term:
x(y^2+1)dx+y(x^2+1)dy=0
Subtracting y(x^2+1)dy on both sides:
x(y^2+1)dx=-y(x^2+1)dy
Divide both sides by (y^2+1)(x^2+1):
x/(x^+2+1)dx=-y/(y^2+1)dy
Integrate both sides..
Use substitution u=x^2+1 and v=y^2+1...so du= 2x dx and dv=2y dy.
.5du/u=-.5dv/v
.5ln|u|=-.5ln|v|+c
Multiply both sides by 2:
ln|u|=-ln|v|+c
Plug in our substitutions :
ln|x^2+1|=-ln|y^2+1|+c
Absolute values unnecessary since insides are positive always.... also -ln(y^2+1)=ln(1/(y^2+1))
e^ on both sides:
x^2+1=k(1/y^2+1)
(y^2+1)(x^2+1)=k
The equation 2 has a graph which is a straight line.
Why?
We can know which of the given equations has a graph which is a straight line just checking the exponents of the variables.
We must remember that every variable that has an exponent equal or higher than 2 (quadratic) will not have a straight line as a graphic.
So, checking the exponents from the given equations, we have:
Hence, we can see that the only equation that has a linear term (straight line graph), is the second equation.
Have a nice day!
Note: I have attached a image for better understanding.
The answer to this question is B
Subtract 9 from both sides of the '=' sign
9 - 9 + b = -8 -9
b = -17 That's the answer.
