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Solve the differential equation $\left(2x-1\right)dx+\left(3y+7\right)dy=0$

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 Final answer to the problem

$\frac{3}{2}y^2+7y=C_0-x^2+x$
Got another answer? Verify it here!

 Step-by-step Solution 

How should I solve this problem?

• Choose an option
• Exact Differential Equation
• Linear Differential Equation
• Separable Differential Equation
• Homogeneous Differential Equation
• Integrate by partial fractions
• Product of Binomials with Common Term
• FOIL Method
• Integrate by substitution
• Integrate by parts
Can't find a method? Tell us so we can add it.
1

The differential equation $\left(2x-1\right)dx+\left(3y+7\right)dy=0$ 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 $f(x,y)=C$

$\left(2x-1\right)dx+\left(3y+7\right)dy=0$
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Using the test for exactness, we check that the differential equation is exact

$0=0$
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Integrate $M(x,y)$ with respect to $x$ to get

$x^2-x+g(y)$
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Now take the partial derivative of $x^2-x$ with respect to $y$ to get

$0+g'(y)$
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Set $3y+7$ and $0+g'(y)$ equal to each other and isolate $g'(y)$

$g'(y)=3y+7$
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Find $g(y)$ integrating both sides

$g(y)=\frac{3}{2}y^2+7y$
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We have found our $f(x,y)$ and it equals

$f(x,y)=x^2-x+\frac{3}{2}y^2+7y$
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Then, the solution to the differential equation is

$x^2-x+\frac{3}{2}y^2+7y=C_0$
9

Group the terms of the equation

$\frac{3}{2}y^2+7y=C_0-x^2+x$

 Final answer to the problem

$\frac{3}{2}y^2+7y=C_0-x^2+x$

 Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

SnapXam A2

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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: Implicit Differentiation

Implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. For differentiating an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y(x) and then differentiate. Instead, one can differentiate R(x, y) with respect to x and y and then solve a linear equation in dy/dx for getting explicitly the derivative in terms of x and y.