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Algebra Baldor
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Solve the differential equation $y^{\prime}+\sqrt{1+x^2}=0$

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Solution

Formulas

Videos

Basic Integrals

· Integral of a Constant
$\int cdx=cvar+C$
· Constant factor Rule
$\int cxdx=c\int xdx$

Trigonometric Integrals

$\int\sec\left(\theta \right)^ndx=\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{\left(n-1\right)}}{n-1}+\frac{n-2}{n-1}\int\sec\left(\theta \right)^{\left(n-2\right)}dx$
$\int\sec\left(\theta \right)dx=\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)+C$

Function Plot

Plotting: $y=\frac{-x\sqrt{1+x^2}}{2}-\frac{1}{2}\ln\left(\sqrt{1+x^2}+x\right)+C_0$

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0
a
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d
f
g
m
n
u
v
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x
y
z
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(◻)
+
-
×
◻/◻
/
÷
◻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

How to improve your answer:

Similar Problems Solved

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

$\frac{dy}{dx}=y^2-9$

$\frac{dy}{dx}=y^2-4$

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

$\frac{dy}{dx}=y\left(y+2\right)$

$\frac{dy}{dx}=\left(1+e^{-x}\right)\left(y^2-1\right)$

$\frac{dy}{dx}=\frac{x}{2x-y}$

Main Topic: Integration by Trigonometric Substitution

Trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities to simplify certain integrals containing radical expressions.

Used Formulas

See formulas (4)

Related Topics

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