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Solve the differential equation $\left(x+y\right)dx+x\cdot dy=0$

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$y=\frac{C_0}{x}-\frac{1}{2}x$
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 Step-by-step Solution 

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The differential equation $\left(x+y\right)dx+x\cdot dy=0$ 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 $f(x,y)=C$

$\left(x+y\right)dx+x\cdot dy=0$

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

$\left(x+y\right)dx+x\cdot dy=0$

Learn how to solve differential equations problems step by step online. Solve the differential equation (x+y)dx+xdy=0. The differential equation \left(x+y\right)dx+x\cdot dy=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(x,y) with respect to x to get. Now take the partial derivative of \frac{1}{2}x^2+yx with respect to y to get.

$y=\frac{C_0}{x}-\frac{1}{2}x$

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Linear Differential EquationExact Differential EquationSeparable Differential EquationHomogeneous Differential Equation

SnapXam A2

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3
4
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

Main Topic: Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives.

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