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Solve the differential equation $e^{\left(x+1\right)}\sin\left(x\right)\cdot dx+\left(2y+1\right)e^{-y^2}dy=0$

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Solution

Final answer to the problem

$\frac{-2+\sqrt{\pi }e^{\left(y^2\right)}\mathrm{erf}\left(y\right)}{2e^{\left(y^2\right)}}=\frac{1}{2}\left(-e^{\left(x+1\right)}\sin\left(x\right)+e^{\left(x+1\right)}\cos\left(x\right)\right)$
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Step-by-step Solution

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  • 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
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1

Grouping the terms of the differential equation

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$\left(2y+1\right)e^{-y^2}dy=-e^{\left(x+1\right)}\sin\left(x\right)\cdot dx$

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Unlock the first 3 steps of this solution

Learn how to solve problems step by step online. Solve the differential equation e^(x+1)sin(x)dx+(2y+1)e^(-y^2)dy=0. Grouping the terms of the differential equation. Integrate both sides of the differential equation, the left side with respect to y, and the right side with respect to x. Solve the integral \int\left(2y+1\right)e^{-y^2}dy and replace the result in the differential equation. The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors.

Final answer to the problem

$\frac{-2+\sqrt{\pi }e^{\left(y^2\right)}\mathrm{erf}\left(y\right)}{2e^{\left(y^2\right)}}=\frac{1}{2}\left(-e^{\left(x+1\right)}\sin\left(x\right)+e^{\left(x+1\right)}\cos\left(x\right)\right)$

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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

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Function Plot

Plotting: $e^{\left(x+1\right)}\sin\left(x\right)\cdot dx+\left(2y+1\right)e^{-y^2}dy$

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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
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

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