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$\frac{dy}{dx}=y^2-1$

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Solution

Final answer to the problem

$-\frac{1}{2}\ln\left|y+1\right|+\frac{1}{2}\ln\left|y-1\right|=x+C_0$
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Step-by-step Solution

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  • Scegliere un'opzione
  • Equazione differenziale esatta
  • Equazione differenziale lineare
  • Equazione differenziale separabile
  • Equazione differenziale omogenea
  • Prodotto di binomi con termine comune
  • Metodo FOIL
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1

Group the terms of the differential equation. Move the terms of the $y$ variable to the left side, and the terms of the $x$ variable to the right side of the equality

$\frac{1}{y^2-1}dy=dx$

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$\frac{1}{y^2-1}dy=dx$

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Learn how to solve problems step by step online. dy/dx=y^2-1. Group the terms of the differential equation. Move the terms of the y variable to the left side, and the terms of the x variable to the right side of the equality. Apply the formula: b\cdot dy=dx\to \int bdy=\int1dx, where b=\frac{1}{y^2-1}. Solve the integral \int\frac{1}{y^2-1}dy and replace the result in the differential equation. Solve the integral \int1dx and replace the result in the differential equation.

Final answer to the problem

$-\frac{1}{2}\ln\left|y+1\right|+\frac{1}{2}\ln\left|y-1\right|=x+C_0$

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Function Plot

Plotting: $\frac{dy}{dx}-y^2+1$

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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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