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Sagot :
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
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