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Solve the differential equation $\frac{dy}{dx}+e^{\left(x-2y\right)}=0$

Step-by-step Solution

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Solution

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Final answer to the problem

$y=\frac{\ln\left(-2e^x+C_1\right)}{2}$
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Step-by-step Solution

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Learn how to solve integration by substitution problems step by step online. Solve the differential equation dy/dx+e^(x-2y)=0. Grouping the terms of the differential equation. x+0=x, where x is any expression. Apply the property of the product of two powers of the same base in reverse: a^{m+n}=a^m\cdot a^n. 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.

Final answer to the problem

$y=\frac{\ln\left(-2e^x+C_1\right)}{2}$

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Function Plot

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

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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
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

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$ydy=4x\left(y^2+1\right)^{\frac{1}{2}}dx$

$\frac{dy}{dx}=\sin5x$

$\frac{dy}{dx}=\frac{x^2-y^2}{3xy}$

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

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

Main Topic: Integration by Substitution

In calculus, integration by substitution, also known as u-substitution, is a method for finding integrals. Using the fundamental theorem of calculus often requires finding an antiderivative.

Used Formulas

See formulas (3)

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