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Here, we show you a step-by-step solved example of separable differential equations. This solution was automatically generated by our smart calculator:
(2xy−y)dx+(x2+x)dy=0
2
Factoring by y
y(2x−1)dx+(x2+x)dy=0
3
Grouping the terms of the differential equation
(x2+x)dy=−(2x−1)y⋅dx
4
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
y1dy=x2+x−(2x−1)dx
Zwischenschritte
Factor the polynomial x2+x by it's greatest common factor (GCF): x
x(x+1)−(2x−1)dx
5
Simplify the expression x2+x−(2x−1)dx
y1dy=x(x+1)−(2x−1)dx
6
Integrate both sides of the differential equation, the left side with respect to y, and the right side with respect to x
∫y1dy=∫x(x+1)−(2x−1)dx
Zwischenschritte
The integral of the inverse of the lineal function is given by the following formula, ∫x1dx=ln(x)
ln∣y∣
7
Solve the integral ∫y1dy and replace the result in the differential equation
ln∣y∣=∫x(x+1)−(2x−1)dx
Zwischenschritte
Take out the constant −1 from the integral
−∫x(x+1)2x−1dx
Rewrite the fraction x(x+1)2x−1 in 2 simpler fractions using partial fraction decomposition
x−1+x+13
Expand the integral ∫(x−1+x+13)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately
−∫x−1dx−∫x+13dx
We can solve the integral ∫x+13dx by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it u), which when substituted makes the integral easier. We see that x+1 it's a good candidate for substitution. Let's define a variable u and assign it to the choosen part
u=x+1
Now, in order to rewrite dx in terms of du, we need to find the derivative of u. We need to calculate du, we can do that by finding the derivative of the equation above
du=dx
Substituting u and dx in the integral and simplify
−∫x−1dx−∫u3du
The integral of the inverse of the lineal function is given by the following formula, ∫x1dx=ln(x)
−−ln∣x∣−∫u3du
Multiply −1 times −1
ln∣x∣−∫u3du
The integral of the inverse of the lineal function is given by the following formula, ∫x1dx=ln(x)
ln∣x∣−3ln∣u∣
Multiply −1 times 3
ln∣x∣−3ln∣u∣
Replace u with the value that we assigned to it in the beginning: x+1
ln∣x∣−3ln∣x+1∣
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration C
ln∣x∣−3ln∣x+1∣+C0
8
Solve the integral ∫x(x+1)−(2x−1)dx and replace the result in the differential equation
ln∣y∣=ln∣x∣−3ln∣x+1∣+C0
Zwischenschritte
Take the variable outside of the logarithm
eln(y)=e(ln(x)−3ln(x+1)+C0)
Simplifying the logarithm
y=e(ln(x)−3ln(x+1)+C0)
Simplify e(ln(x)−3ln(x+1)+C0) by applying the properties of exponents and logarithms
y=xe(−3ln(x+1)+C0)
Simplify e(−3ln(x+1)+C0) by applying the properties of exponents and logarithms
y=eC0x(x+1)−3
We can rename eC0 as other constant
y=C1x(x+1)−3
Applying the property of exponents, a−n=an1, where n is a number
y=C1x(x+1)31
Multiply the fraction by the term
y=(x+1)3C1x
9
Find the explicit solution to the differential equation. We need to isolate the variable y
y=(x+1)3C1x
Endgültige Antwort auf das Problem
y=(x+1)3C1x
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