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Homogeneous Differential Equation Calculator

Get detailed solutions to your math problems with our Homogeneous Differential Equation step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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Example

Solved Problems

Difficult Problems

1

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

$\frac{dy}{dx}=\frac{x^2+y^2}{xy}$
2

We can identify that the differential equation $\frac{dy}{dx}=\frac{x^2+y^2}{xy}$ is homogeneous, since it is written in the standard form $\frac{dy}{dx}=\frac{M(x,y)}{N(x,y)}$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and both are homogeneous functions of the same degree

$\frac{dy}{dx}=\frac{x^2+y^2}{xy}$
3

Use the substitution: $y=ux$

$\frac{u\cdot dx+x\cdot du}{dx}=\frac{x^2+\left(ux\right)^2}{xux}$

When multiplying two powers that have the same base ($x$), you can add the exponents

$\frac{u\cdot dx+x\cdot du}{dx}=\frac{x^2+\left(ux\right)^2}{x^2u}$

The power of a product is equal to the product of it's factors raised to the same power

$\frac{u\cdot dx+x\cdot du}{dx}=\frac{x^2+u^2x^2}{x^2u}$

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

$\frac{u\cdot dx+x\cdot du}{dx}=\frac{x^2\left(1+u^2\right)}{x^2u}$

Simplify the fraction

$\frac{u\cdot dx+x\cdot du}{dx}=\frac{1+u^2}{u}$

Expand the fraction $\frac{u\cdot dx+x\cdot du}{dx}$ into $2$ simpler fractions with common denominator $dx$

$\frac{u\cdot dx}{dx}+\frac{x\cdot du}{dx}=\frac{1+u^2}{u}$

Simplify the resulting fractions

$u+\frac{x\cdot du}{dx}=\frac{1+u^2}{u}$

We need to isolate the dependent variable $u$, we can do that by simultaneously subtracting $u$ from both sides of the equation

$\frac{x\cdot du}{dx}=\frac{1+u^2}{u}-u$

Combine all terms into a single fraction with $u$ as common denominator

$\frac{x\cdot du}{dx}=\frac{1+u^2-u^2}{u}$

Cancel like terms $u^2$ and $-u^2$

$\frac{x\cdot du}{dx}=\frac{1}{u}$

Group the terms of the differential equation. Move the terms of the $u$ variable to the left side, and the terms of the $x$ variable to the right side of the equality

$u\cdot du=\frac{1}{x}dx$
4

Expand and simplify

$u\cdot du=\frac{1}{x}dx$
5

Integrate both sides of the differential equation, the left side with respect to $u$, and the right side with respect to $x$

$\int udu=\int\frac{1}{x}dx$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$\frac{1}{2}u^2$
6

Solve the integral $\int udu$ and replace the result in the differential equation

$\frac{1}{2}u^2=\int\frac{1}{x}dx$

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

$\ln\left|x\right|$

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

$\ln\left|x\right|+C_0$
7

Solve the integral $\int\frac{1}{x}dx$ and replace the result in the differential equation

$\frac{1}{2}u^2=\ln\left|x\right|+C_0$
8

Replace $u$ with the value $\frac{y}{x}$

$\frac{1}{2}\left(\frac{y}{x}\right)^2=\ln\left(x\right)+C_0$
9

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{1}{2}\frac{y^2}{x^2}=\ln\left|x\right|+C_0$

Multiplying fractions $\frac{1}{2} \times \frac{y^2}{x^2}$

$\frac{y^2}{2x^2}$
10

Multiplying fractions $\frac{1}{2} \times \frac{y^2}{x^2}$

$\frac{y^2}{2x^2}=\ln\left|x\right|+C_0$

Multiply both sides of the equation by $2x^2$

$y^2=2\left(\ln\left(x\right)+C_0\right)x^2$

Removing the variable's exponent

$\sqrt{y^2}=\pm \sqrt{2\left(\ln\left(x\right)+C_0\right)x^2}$

Cancel exponents $2$ and $1$

$y=\pm \sqrt{2\left(\ln\left(x\right)+C_0\right)x^2}$

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$y=\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2},\:y=-\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2}$

Combining all solutions, the $2$ solutions of the equation are

$y=\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2},\:y=-\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2}$
11

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

$y=\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2},\:y=-\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2}$

Final answer to the problem

$y=\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2},\:y=-\sqrt{2\left(\ln\left(x\right)+C_0\right)x^2}$

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