Solve the differential equation $\frac{dy}{dx}=\tan\left(x+y\right)$

Step-by-step Solution

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Final answer to the problem

$\ln\left(\frac{\sqrt{-\left(\tan\left(\frac{x+y}{2}\right)-1\right)^2+2}}{\sqrt{2}}\right)-\frac{1}{2}\ln\left(1+\tan\left(\frac{x+y}{2}\right)^{2}\right)+\arctan\left(\tan\left(\frac{x+y}{2}\right)\right)=x+C_0$
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Step-by-step Solution

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When we identify that a differential equation has an expression of the form $Ax+By+C$, we can apply a linear substitution in order to simplify it to a separable equation. We can identify that $x+y$ has the form $Ax+By+C$. Let's define a new variable $u$ and set it equal to the expression

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$u=x+y$

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Learn how to solve weierstrass substitution problems step by step online. Solve the differential equation dy/dx=tan(x+y). When we identify that a differential equation has an expression of the form Ax+By+C, we can apply a linear substitution in order to simplify it to a separable equation. We can identify that x+y has the form Ax+By+C. Let's define a new variable u and set it equal to the expression. Isolate the dependent variable y. Differentiate both sides of the equation with respect to the independent variable x. Now, substitute x+y and \frac{dy}{dx} on the original differential equation. We will see that it results in a separable equation that we can easily solve.

Final answer to the problem

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

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Function Plot

Plotting: $\frac{dy}{dx}-\tan\left(x+y\right)$

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

e
π
ln
log
log
lim
d/dx
Dx
|◻|
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:

Main Topic: Weierstrass Substitution

In integral calculus, the Weierstrass substitution or tangent half angle substitution is a method for solving integrals, which converts a rational expression of trigonometric functions into an algebraic rational function, which can be easier to integrate. The Weierstrass substitution is very useful for integrals that involve a simple rational expression with sine and/or cosine in the denominator.

Used Formulas

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