Solve the differential equation $\frac{dx}{dp}=\frac{-\left(x-pe^p\right)}{p}$

Step-by-step Solution

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Final answer to the problem

$x=\frac{C_0+e^p\cdot p-e^p}{p}$
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Step-by-step Solution

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Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$

Learn how to solve differential calculus problems step by step online.

$pdx1\left(x-pe^p\right)dp=0$

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Learn how to solve differential calculus problems step by step online. Solve the differential equation dx/dp=(-(x-pe^p))/p. Rewrite the differential equation in the standard form M(x,y)dx+N(x,y)dy=0. The differential equation pdx1\left(x-pe^p\right)dp=0 is exact, since it is written in the standard form M(x,y)dx+N(x,y)dy=0, where M(x,y) and N(x,y) are the partial derivatives of a two-variable function f(x,y) and they satisfy the test for exactness: \displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form f(x,y)=C. Using the test for exactness, we check that the differential equation is exact. Integrate M(p,x) with respect to p to get.

Final answer to the problem

$x=\frac{C_0+e^p\cdot p-e^p}{p}$

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Function Plot

Plotting: $\frac{dx}{dp}+\frac{x-pe^p}{p}$

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5
6
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
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: Differential Calculus

The derivative of a function of a real variable measures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool of calculus.

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