Homogeneous, Exact and Linear Equations

Homogeneous Differential Equations

A first-order differential equation of the form $\frac{\mathrm{dy}}{\mathrm{dx}}=f(x,y)$ is called homogeneous if the function $f(x,y)$ can be expressed as a function of the ratio $\frac{y}{x}.$ This means that $f(\mathrm{kx},\mathrm{ky})=f(x,y)$ for any non-zero constant $k.$ Alternatively, a differential equation can be written in the form $M(x,y)\mathrm{dx}+N(x,y)\mathrm{dy}=0$ where $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree. A function $F(x,y)$ is homogeneous of degree $n$ if $F(\mathrm{kx},\mathrm{ky})=k^{n}F(x,y)$.

Solving Homogeneous Equations

To solve a homogeneous differential equation, we use the substitution $y=\mathrm{vx}.$ Differentiating this with respect to $x$, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}}=v+x\frac{\mathrm{dv}}{\mathrm{dx}}$

Substitute $y=\mathrm{vx}$ and the expression for $\frac{\mathrm{dy}}{\mathrm{dx}}$ into the original differential equation. This will transform the equation into a separable equation in terms of $v$ and $x$.

The steps are:

1. Identify if the equation is homogeneous. Check if $f(x,y)$ depends only on $\frac{y}{x}$, or if $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree.
2. Make the substitution $y=\mathrm{vx}$, which implies $\frac{\mathrm{dy}}{\mathrm{dx}}=v+x\frac{\mathrm{dv}}{\mathrm{dx}}$.
3. Substitute these into the differential equation to get an equation in terms of $v$ and $x$.
4. Separate the variables $v$ and $x$.
5. Integrate both sides.
6. Substitute back $v=\frac{y}{x}$ to get the general solution in terms of $x$ and $y$.

Example 1:

Solve the differential equation $(x^2+y^2)\mathrm{dx}-2\mathrm{xy}\mathrm{dy}=0$.

This can be written as $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{x^2+y^2}{2\mathrm{xy}}=\frac{1+(\frac{y}{x}{)}^2}{2(\frac{y}{x})}$. This is a homogeneous equation.

Let $y=\mathrm{vx}$, so $\frac{\mathrm{dy}}{\mathrm{dx}}=v+x\frac{\mathrm{dv}}{\mathrm{dx}}$.

Substituting these into the equation: $v+x\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+v^2}{2v}$

$x\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+v^2}{2v}-v=\frac{1+v^2-2v^2}{2v}=\frac{1-v^2}{2v}$

Separating variables: $\frac{2v}{1-v^2}\mathrm{dv}=\frac{\mathrm{dx}}{x}$

Integrate both sides: $\int \frac{2v}{1-v^2}\mathrm{dv}=\int \frac{\mathrm{dx}}{x}$

Let $u=1-v^2,\mathrm{du}=-2v\mathrm{dv}$. $\int -\frac{\mathrm{du}}{u}=\int \frac{\mathrm{dx}}{x}$

$-\ln |u|=\ln |x|+C$

$-\ln |1-v^2|=\ln |x|+C$

$\ln |1-v^2{|}^{-1}=\ln |x|+C$

$\ln |\frac{1}{1-v^2}|=\ln |x|+C$

$\frac{1}{1-v^2}=Ax$, where $A=\pm e^C$ is an arbitrary constant.

Substitute back $v=\frac{y}{x}$: $\frac{1}{1-\frac{y^2}{x^2}}=Ax$

$\frac{x^2}{x^2-y^2}=Ax$

$x^2=Ax(x^2-y^2)$

$x=A(x^2-y^2)$

This is the general solution.

Exact Differential Equations

A first-order differential equation of the form $M(x,y)\mathrm{dx}+N(x,y)\mathrm{dy}=0$ is called an exact differential equation if there exists a function $F(x,y)$ such that its total differential $\mathrm{dF}$ is equal to the left side of the equation. That is, $\mathrm{dF}=\frac{\partial F}{\partial x}\mathrm{dx}+\frac{\partial F}{\partial y}\mathrm{dy}=M(x,y)\mathrm{dx}+N(x,y)\mathrm{dy}$. This implies that $M=\frac{\partial F}{\partial x}$ and $N=\frac{\partial F}{\partial y}$.

Condition for Exactness

A differential equation $M(x,y)\mathrm{dx}+N(x,y)\mathrm{dy}=0$ is exact if and only if: $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ This condition arises from Clairaut's Theorem (or Schwarz's Theorem) on the equality of mixed partial derivatives: $\frac{{\partial }^{2}F}{\partial y\partial x}=\frac{{\partial }^{2}F}{\partial x\partial y}$.

Solving Exact Equations

If the condition for exactness is met, the general solution is given by $F(x,y)=C$, where $C$ is an arbitrary constant. We can find $F(x,y)$ using the following steps:

1. Check if the equation is exact by verifying $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.
2. Integrate $M$ with respect to $x$, treating $y$ as a constant. This gives a function $F(x,y)=\int M(x,y)\mathrm{dx}+g(y)$, where $g(y)$ is an arbitrary function of $y$ (analogous to the constant of integration).
3. Differentiate this expression for $F(x,y)$ with respect to $y$: $\frac{\partial F}{\partial y}=\frac{\partial }{\partial }(\int M(x,y)\mathrm{dx})+\mathrm{g\text{'}}(y)$
4. Equate this to $N$: $\frac{\partial F}{\partial y}=N(x,y)$
5. Solve for $\mathrm{g\text{'}}(y)$. You should find that the terms involving $x$ cancel out, leaving an expression solely in terms of $y$.
6. Integrate $\mathrm{g\text{'}}(y)$ with respect to $y$ to find $g(y)$.
7. Substitute $g(y)$ back into the expression for $F(x,y)$.
8. The general solution is $F(x,y)=C$.

Alternatively, one can integrate $N$ with respect to $y$ first, yielding $F(x,y)=\int N(x,y)\mathrm{dy}+h(x)$, and then proceed similarly.

Example 2:

Solve the differential equation $(2x^{2}y-3)\mathrm{dx}+(2x^3+4y)\mathrm{dy}=0$.

Here, $M(x,y)=2x^{2}y-3$ and $N(x,y)=2x^3+4y$.

Check for exactness: $\frac{\partial M}{\partial y}=\frac{\partial }{\partial }(2x^{2}y-3)=2x^2$

$\frac{\partial N}{\partial x}=\frac{\partial }{\partial }(2x^3+4y)=6x^2$

Since $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$, the equation is not exact as given.

Let's re-examine the problem statement. It seems there might be a typo in the original problem. Let's assume the equation was intended to be exact. If we assume $N(x,y)=x^3+4y$, then $\frac{\partial N}{\partial x}=3x^2$, still not equal.

Let's assume the equation was: $(2x^{2}y-3)\mathrm{dx}+(x^3+4y)\mathrm{dy}=0$. Here $M=2x^{2}y-3$, $N=x^3+4y$. $\frac{\partial M}{\partial y}=2x^2$ $\frac{\partial N}{\partial x}=3x^2$ Still not exact.

Let's try another common form for exact equations for illustration. Solve $(x^2+y^2)\mathrm{dx}+2\mathrm{xy}\mathrm{dy}=0$. $M=x^2+y^2$, $N=2\mathrm{xy}$. $\frac{\partial M}{\partial y}=2y$ $\frac{\partial N}{\partial x}=2y$ Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, the equation is exact.

Using the shortcut: $\int M\mathrm{dx}+\int (\text{terms in}N\text{free of}x)\mathrm{dy}=C$

$\int (x^2+y^2)\mathrm{dx}+\int (\text{no terms free of x in N})\mathrm{dy}=C$

$(\frac{x^3}{3}+x^{}{y}^2)+0=C$

The solution is $\frac{x^3}{3}+x{y}^2=C$.

Linear Differential Equations

A first-order differential equation is called linear if it can be written in the standard form: $\frac{\mathrm{dy}}{\mathrm{dx}}+P(x)y=Q(x)$ where $P(x)$ and $Q(x)$ are functions of $x$ only, or constants.

The term $P(x)y$ is linear in $y$, and the highest derivative is the first derivative.

Solving Linear Equations using an Integrating Factor

The method to solve a linear differential equation involves using an integrating factor. The integrating factor, denoted by $\mathrm{I.F.}$, is given by: $\mathrm{I.F.}=e^{\int P(x)\mathrm{dx}}$

Multiplying the standard form of the linear equation by the integrating factor $\mathrm{I.F.}$, the left side becomes the derivative of the product of the integrating factor and $y$: $\mathrm{I.F.}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{I.F.}P(x)y=\mathrm{I.F.}Q(x)$ $\frac{d}{\mathrm{dx}}(y\cdot \mathrm{I.F.})=\mathrm{I.F.}Q(x)$

Now, integrate both sides with respect to $x$: $\int \frac{d}{\mathrm{dx}}(y\cdot \mathrm{I.F.})\mathrm{dx}=\int \mathrm{I.F.}Q(x)\mathrm{dx}$ $y\cdot \mathrm{I.F.}=\int \mathrm{I.F.}Q(x)\mathrm{dx}+C$

Finally, solve for $y$: $y=\frac{1}{\mathrm{I.F.}}(\int \mathrm{I.F.}Q(x)\mathrm{dx}+C)$

Example 3:

Solve the differential equation $x\frac{\mathrm{dy}}{\mathrm{dx}}+y=x^2$.

First, rewrite the equation in standard form: $\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{x}y=x$

Here, $P(x)=\frac{1}{x}$ and $Q(x)=x$.

Calculate the integrating factor: $\int P(x)\mathrm{dx}=\int \frac{1}{x}\mathrm{dx}=\ln |x|$ $\mathrm{I.F.}=e^{\ln |x|}=|x|$ For simplicity, we can take $\mathrm{I.F.}=x$ (assuming $x>0$).

Multiply the standard form by $x$: $x\frac{\mathrm{dy}}{\mathrm{dx}}+x(\frac{1}{x})y=x\cdot x$ $x\frac{\mathrm{dy}}{\mathrm{dx}}+y=x^2$

The left side is the derivative of $y\cdot x$: $\frac{d}{\mathrm{dx}}(yx)=x^2$

Integrate both sides with respect to $x$: $\int \frac{d}{\mathrm{dx}}(yx)\mathrm{dx}=\int x^2\mathrm{dx}$ $yx=\frac{x^3}{3}+C$

Solve for $y$: $y=\frac{x^3}{3x}+\frac{C}{x}$ $y=\frac{x^2}{3}+\frac{C}{x}$

This is the general solution.

Example 4: Bernoulli Equation (Can be reduced to Linear)

A Bernoulli equation has the form: $\frac{\mathrm{dy}}{\mathrm{dx}}+P(x)y=Q(x)y^n$ where $n$ is any real number, and typically $n\ne 0$, $n\ne 1$.

To solve a Bernoulli equation, we use the substitution $v=y^{1-n}$. This substitution transforms the Bernoulli equation into a linear differential equation in terms of $v$.

Let's solve $x\frac{\mathrm{dy}}{\mathrm{dx}}+y=x{y}^2$. This is a Bernoulli equation with $P(x)=\frac{1}{x}$, $Q(x)=1$, and $n=2$.

The substitution is $v=y^{1-2}=y^{-1}=\frac{1}{y}$. Then $\frac{\mathrm{dv}}{\mathrm{dx}}=-1y^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{y^2}\frac{\mathrm{dy}}{\mathrm{dx}}$.

Divide the original equation by $x$: $\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{x}y=y{y}^2$

Multiply by $-\frac{1}{y^2}$: $-\frac{1}{y^2}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{x}\frac{1}{y}=-1$

Substitute $v=\frac{1}{y}$ and $\frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{1}{y^2}\frac{\mathrm{dy}}{\mathrm{dx}}$: $\frac{\mathrm{dv}}{\mathrm{dx}}-\frac{1}{x}v=-1$

This is a linear equation. $P(x)=-\frac{1}{x}$, $Q(x)=-1$.

Integrating factor: $\mathrm{I.F.}=e^{\int (-\frac{1}{x})\mathrm{dx}}=e^{-\ln |x|}=e^{\ln |x{|}^{-1}}=\frac{1}{|x|}$. Let's use $\mathrm{I.F.}=\frac{1}{x}$.

Multiply the linear equation by $\frac{1}{x}$: $\frac{1}{x}\frac{\mathrm{dv}}{\mathrm{dx}}-\frac{1}{x^2}v=-\frac{1}{x}$

The left side is $\frac{d}{\mathrm{dx}}(v\cdot \frac{1}{x})$. $\frac{d}{\mathrm{dx}}(\frac{v}{x})=-\frac{1}{x}$

Integrate both sides: $\int \frac{d}{\mathrm{dx}}(\frac{v}{x})\mathrm{dx}=\int -\frac{1}{x}\mathrm{dx}$ $\frac{v}{x}=-\ln |x|+C$

Solve for $v$: $v=-x\ln |x|+\mathrm{Cx}$

Substitute back $v=\frac{1}{y}$: $\frac{1}{y}=-x\ln |x|+\mathrm{Cx}$

The general solution for $y$ is: $y=\frac{1}{\mathrm{Cx}-x\ln |x|}$