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Trennbare Differentialgleichungen Rechner

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1

Here, we show you a step-by-step solved example of separable differential equations. This solution was automatically generated by our smart calculator:

(2xyy)dx+(x2+x)dy=0\left(2xy-y\right)dx+\left(x^2+x\right)dy=0
2

Factoring by yy

y(2x1)dx+(x2+x)dy=0y\left(2x-1\right)dx+\left(x^2+x\right)dy=0
3

Grouping the terms of the differential equation

(x2+x)dy=(2x1)ydx\left(x^2+x\right)dy=-\left(2x-1\right)y\cdot dx
4

Group the terms of the differential equation. Move the terms of the yy variable to the left side, and the terms of the xx variable to the right side of the equality

1ydy=(2x1)x2+xdx\frac{1}{y}dy=\frac{-\left(2x-1\right)}{x^2+x}dx

Factor the polynomial x2+xx^2+x by it's greatest common factor (GCF): xx

(2x1)x(x+1)dx\frac{-\left(2x-1\right)}{x\left(x+1\right)}dx
5

Simplify the expression (2x1)x2+xdx\frac{-\left(2x-1\right)}{x^2+x}dx

1ydy=(2x1)x(x+1)dx\frac{1}{y}dy=\frac{-\left(2x-1\right)}{x\left(x+1\right)}dx
6

Integrate both sides of the differential equation, the left side with respect to yy, and the right side with respect to xx

1ydy=(2x1)x(x+1)dx\int\frac{1}{y}dy=\int\frac{-\left(2x-1\right)}{x\left(x+1\right)}dx

The integral of the inverse of the lineal function is given by the following formula, 1xdx=ln(x)\displaystyle\int\frac{1}{x}dx=\ln(x)

lny\ln\left|y\right|
7

Solve the integral 1ydy\int\frac{1}{y}dy and replace the result in the differential equation

lny=(2x1)x(x+1)dx\ln\left|y\right|=\int\frac{-\left(2x-1\right)}{x\left(x+1\right)}dx

Take out the constant 1-1 from the integral

2x1x(x+1)dx-\int\frac{2x-1}{x\left(x+1\right)}dx

Rewrite the fraction 2x1x(x+1)\frac{2x-1}{x\left(x+1\right)} in 22 simpler fractions using partial fraction decomposition

1x+3x+1\frac{-1}{x}+\frac{3}{x+1}

Expand the integral (1x+3x+1)dx\int\left(\frac{-1}{x}+\frac{3}{x+1}\right)dx into 22 integrals using the sum rule for integrals, to then solve each integral separately

1xdx3x+1dx-\int\frac{-1}{x}dx-\int\frac{3}{x+1}dx

We can solve the integral 3x+1dx\int\frac{3}{x+1}dx 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 uu), which when substituted makes the integral easier. We see that x+1x+1 it's a good candidate for substitution. Let's define a variable uu and assign it to the choosen part

u=x+1u=x+1

Now, in order to rewrite dxdx in terms of dudu, we need to find the derivative of uu. We need to calculate dudu, we can do that by finding the derivative of the equation above

du=dxdu=dx

Substituting uu and dxdx in the integral and simplify

1xdx3udu-\int\frac{-1}{x}dx-\int\frac{3}{u}du

The integral of the inverse of the lineal function is given by the following formula, 1xdx=ln(x)\displaystyle\int\frac{1}{x}dx=\ln(x)

lnx3udu- -\ln\left|x\right|-\int\frac{3}{u}du

Multiply 1-1 times 1-1

lnx3udu\ln\left|x\right|-\int\frac{3}{u}du

The integral of the inverse of the lineal function is given by the following formula, 1xdx=ln(x)\displaystyle\int\frac{1}{x}dx=\ln(x)

lnx3lnu\ln\left|x\right|- 3\ln\left|u\right|

Multiply 1-1 times 33

lnx3lnu\ln\left|x\right|-3\ln\left|u\right|

Replace uu with the value that we assigned to it in the beginning: x+1x+1

lnx3lnx+1\ln\left|x\right|-3\ln\left|x+1\right|

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration CC

lnx3lnx+1+C0\ln\left|x\right|-3\ln\left|x+1\right|+C_0
8

Solve the integral (2x1)x(x+1)dx\int\frac{-\left(2x-1\right)}{x\left(x+1\right)}dx and replace the result in the differential equation

lny=lnx3lnx+1+C0\ln\left|y\right|=\ln\left|x\right|-3\ln\left|x+1\right|+C_0

Take the variable outside of the logarithm

eln(y)=e(ln(x)3ln(x+1)+C0)e^{\ln\left(y\right)}=e^{\left(\ln\left(x\right)-3\ln\left(x+1\right)+C_0\right)}

Simplifying the logarithm

y=e(ln(x)3ln(x+1)+C0)y=e^{\left(\ln\left(x\right)-3\ln\left(x+1\right)+C_0\right)}

Simplify e(ln(x)3ln(x+1)+C0)e^{\left(\ln\left(x\right)-3\ln\left(x+1\right)+C_0\right)} by applying the properties of exponents and logarithms

y=xe(3ln(x+1)+C0)y=xe^{\left(-3\ln\left(x+1\right)+C_0\right)}

Simplify e(3ln(x+1)+C0)e^{\left(-3\ln\left(x+1\right)+C_0\right)} by applying the properties of exponents and logarithms

y=eC0x(x+1)3y=e^{C_0}x\left(x+1\right)^{-3}

We can rename eC0e^{C_0} as other constant

y=C1x(x+1)3y=C_1x\left(x+1\right)^{-3}

Applying the property of exponents, an=1an\displaystyle a^{-n}=\frac{1}{a^n}, where nn is a number

y=C1x1(x+1)3y=C_1x\frac{1}{\left(x+1\right)^{3}}

Multiply the fraction by the term

y=C1x(x+1)3y=\frac{C_1x}{\left(x+1\right)^{3}}
9

Find the explicit solution to the differential equation. We need to isolate the variable yy

y=C1x(x+1)3y=\frac{C_1x}{\left(x+1\right)^{3}}

Endgültige Antwort auf das Problem

y=C1x(x+1)3y=\frac{C_1x}{\left(x+1\right)^{3}}

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