Differential Equations -- Important Stuff -- Mathematics III

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Index

1. Basic types of differential equations.
2. 1. Linear Differential equation.
3. 2. Exact equations
4. Steps for finding integrating factor for a non-exact equation.
5. 3. Bernoulli’s equation
6. Differential equations of first order but not first degree.
7. 1. Equations solvable for p
8. 2. Equations solvable for y
9. 3. Equations solvable for x
10. Clairaut’s equation.
11. Second Order Differential Equations
12. Method 1 Using the inverse operator method.
13. Method 2 Using the method of variation of parameters.

Note : If the equations on the website seem a bit to clumped up/clogged, please go to the resources section and download the PDF. It has better indentation.

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Basic types of differential equations.

1. Linear Differential equation.

These can be of two types:

1. $\boxed{\frac{dy}{dx}+P(x).y=Q(x)}$ To solve this, we find an integrating factor(I.F).

$I.F=e^{\int P(x)dx}$

And we find the solution in terms of y as:

$\boxed{y.(I.F)=\int Q(x).(I.F)dx+C}$

2. $\boxed{\frac{dx}{dy}+P(y).x=Q(y)}$ Same stuff, just in terms of x this time.

I.F = $I.F=e^{\int P(y)dy}$

And we find the solution in terms of x as: $\boxed{x.(I.F)=\int Q(y).(I.F)dy+C}$

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2. Exact equations

An ordinary D.E of first order and first degree is of the form: $\frac{dy}{dx}=F(x,y)$ can be written as: $\boxed{Mdx+Ndy=0}$

This equation will be considered exact, ==if and only if==

$\frac{dM}{dy}=\frac{dN}{dx}$ and the solution of the equation will be given by:

$\boxed{y=\int Mdx+\int (no-x-containing-term-from-N)dy=constant}$

However there are times when :

$\frac{dM}{dy}\ne \frac{dN}{dx}$ and thus the equation by-default is not exact.

However we can convert it to an exact equation by finding an integrating factor and then multiplying it to the original equation.

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Steps for finding integrating factor for a non-exact equation.

1. If equation is of the form $\boxed{Mdx+Ndy=0}$
2. Then check if $\frac{dM}{dy}\ne \frac{dN}{dx}$
3. If true, then equation is non-exact.

Proceeding from here:

Case 1

1. Check if both M and N are homogenous (degree of both the functions should be equal to zero).
2. If case 1 is true, then check if both M and N have same degree
3. If true, then check if $Mx+Ny\ne 0$ .
4. If all of the above are true, then the I.F will be: $\boxed{\frac{1}{Mx+Ny}}$
5. Multiply this I.F to the original equation to get the exact equation. Proceed solving from there as previously discussed.

Case 2 (M and N are both not homogenous/M is not homogenous/N is not homogenous)

1. In this case, we proceed by finding : $\frac{dM}{dy}-\frac{dN}{dx}$
2. Now we can either:
   1. Divide by M: If the result is a function of y and y only, then our I.F will be: $\boxed{e^{-\int g(y)dy}}$
   2. Divide by N: If the result is a function of x and x only, then our I.F will be: $\boxed{e^{\int f(x)dx}}$
3. Multiply this I.F to the original equation to get the exact equation. Proceed solving from there as previously discussed.

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For the alternate form

1. If equation is of form $\boxed{y(f(x,y)dx+g(x,y)dy)=0}$
2. M = $y(f(x,y)$ , N = $g(x,y)$.
3. If $Mx-Ny\ne 0$, then
4. Integrating factor = $\boxed{\frac{1}{Mx-Ny}}$
5. Multiply this I.F to the original equation to get the exact equation. Proceed solving from there as previously discussed.

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3. Bernoulli’s equation

It is of the form: $y^{\prime }+P(x)y=Q(x).y^n$

Where $y^{\prime }=\frac{dy}{dx}$ The easiest way to solve this is to reduce to a linear D.E.

We start by dividing both sides of the equation with $y^n$ to get :

$y^{-n}.\frac{dy}{dx}+P(x).y^{1-n}=Q(x)$

Let $y^{1-n}$ = $z$ .

Differentiate both sides w.r.t $x$.

$y^{-n}.\frac{dy}{dx}=\frac{dz}{dx}.\frac{1}{(1-n)}$

Substitute back into the original equation, we get:

$\frac{1}{(1-n)}.\frac{dz}{dx}+P(x).z=Q(x)$

or $\frac{dz}{dx}+(1-n).P(x).z=Q(x).(1-n)$

which can now be solved using the formula of a linear D.E

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Differential equations of first order but not first degree.

1. Equations solvable for p

The required form for this type of equation should in the form of a quadratic equation:

$a{x}^2+bx+c=0$

or

$\boxed{p^2+p+c=0}$

Example explanation

Let’s say we have the equation:

$(\frac{dy}{dx}{)}^2-\frac{dy}{dx}-6=0$

1. Set $\frac{dy}{dx}=p$
2. So the equation becomes $p^2-p-6=0$
3. Solving the equation, we get roots as p = -2 and p = 3.
4. Equation $\frac{dy}{dx}$ to both the values separately and separate the variables

$dy=3dx...(1)$ $dy=-2dx....(2)$

5. Integrate both the equations, we get:

$y=3x+c...(3)$ or $y-3x-c=0$ and $y=-2x+c$ or $y+2x-c=\mathrm{0...}(4)$ 6. Finally we write both the equations in product form as :

$(y-3x-c)(y+2x-c)=0$

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2. Equations solvable for y

These are equations are always of $y^n$ where $n$ is always equal to 1.

Steps to solve an equation for y.

1. Write equation in terms of y, where y is isolated on the LHS.
2. Differentiate L.H.S and R.H.S w.r.t x.
3. Let $\frac{dy}{dx}=p$.
4. Try to solve and reduce to a linear differential equation which will be $\frac{dx}{dp}+P(p).x=Q(p)$, by doing some substitution by setting some coefficient to z.
5. Then solve the linear D.E to get a value in terms of $x$.
6. Put that value back in the original given equation and you have the required solution.

Alternatively .

After step 3

4. Try to solve the equation and get a value of in terms of $p$.
5. Take only those terms which have $\frac{dp}{dx}$ associated with them.
6. Now let’s say you get, $p=bx.\frac{dp}{dx}$, then separate the variables, $x$ to one side, $p$ to the other.
7. Integrate both sides.
8. Put the value of $p$ back in the original equation.

==Both methods are valid. Just depends which one is more applicable, based on the question==.

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Example explanation

Let’s say we have an equation.

$y-2px+x{p}^2=0$

So rewriting the equation we get :

$y=2px-x{p}^2...(1)$ So we differentiate both sides w.r.t $x$.

$\frac{dy}{dx}=2p-p^2+2x.\frac{dp}{dx}[1-p]$

Let $\frac{dy}{dx}=p$.

So we get the equation as :

$p=2p-p^2+2x.\frac{dp}{dx}[1-p]$

which can be further reduced down to:

$(1-p)(p+2x.\frac{dp}{dx})=0$

Here we see that we already have a term with $\frac{dp}{dx}$. So we don’t need to reduce to a linear D.E at this point, we will just consider the second term.

Therefore, $p=-2x.\frac{dp}{dx}$

We separate the variables to get:

$\frac{p}{dp}=\frac{-2x}{dx}$

Now we integrate both sides to get

$2log(p)=-log(x)+log(c)$ which can be solved further down to get:

$log(p^{2}x)=log(c)$

or $p^2=\frac{c}{x}$ or $p=\sqrt{\frac{c}{x}}$

Now we apply this value back in the original equation to get

$y-2\sqrt{\frac{c}{x}}.x-x.\frac{c}{x}=0$

or , $\boxed{y=2\sqrt{c}.\sqrt{x}+c}$

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Example 2.

Let’s say we have another equation: $y=2px-p^2$ We apply the same rules as before to get

$\frac{dy}{dx}=2x.\frac{dp}{dx}+2p-2p.\frac{dp}{dx}$ Let $\frac{dy}{dx}=p$.

and then by solving further we get an equation:

$p+2(x-p).\frac{dp}{dx}=0$

Now we could technically apply the alternative method and only choose the second term with dp/dx, but then separating variables would be a problem in this instance.

We would get $2xdp-2pdp=dx$ Which now becomes troublesome to separate further.

So discarding this approach, we convert this equation into a linear D.E and then try to solve it.

Thus: $p.\frac{dx}{dp}+2x-2p=0$

or $\frac{dx}{dp}+2x=2$

Solving this D.E we get a value in terms of $x$ as :

$x=\frac{2p}{3}+c{p}^{-2}$

Applying this value in original equation, we get the required solution in terms of $y$ as:

$\boxed{y=p^2+2c{p}^{-1}}$

3. Equations solvable for x

Here equations are given in the form of: $x=f(y,p)$ We generally solve this by differentiating both sides w.r.t $y$.

Therefore, $\frac{dx}{dy}=F(y,p.\frac{dp}{dy})$ If $\frac{dy}{dx}=p$ then $\frac{dx}{dy}=\frac{1}{p}$

$\therefore \frac{1}{p}=F(y,p.\frac{dp}{dy})$, which is a first order differential equation in terms of p and y

We will solve the equation further and then separate variables, integrate both sides to get a value in terms of y, put it back in original equation.

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Example explanation

Let’s say we have a given equation : $x=y+p^2$

We differentiate both sides w.r.t $y$.

$\therefore \frac{dx}{dy}=1+2p.\frac{dp}{dy}$

Or, $\frac{1}{p}=1+2p.\frac{dp}{dy}$ Continuing to solve that we get :

$\frac{1-p}{2p^2}=\frac{dp}{dy}$

or

$dy=\frac{2p^2}{1-p}.dp$ or $dy=[2p^2-2p].dp$

Integrating both sides, we got a solution in terms of $y$ as: $y=\frac{2p^3}{3}-p^2+c$

So applying this value in our original equation, we get the required solution as:

$\boxed{x=\frac{2p^3}{3}+c}$

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Clairaut’s equation.

A differential equation of the form

$\boxed{y=px+f(p)}$

is called Clairaut’s form.

We will use the method of 2. Equations solvable for y to solve such equations.

So we differentiate the equation w.r.t $x$.

$\frac{dy}{dx}=\frac{dp}{dx}.[x+f^{\prime }(p)]+p$

Let

$\frac{dy}{dx}=p$

$p=p+\frac{dp}{dx}[x+f^{\prime }(p)]$

or

$\frac{dp}{dx}.[x+f^{\prime }(p)]=0$

Now we set both of the terms

$\frac{dp}{dx}=\mathrm{0...}(1)$

and

$x+f^{\prime }(p)=\mathrm{0....}(2)$

From equation 1, we get $p=c$. Thus,

$y=cx+f(c)$

which is the required general solution,

And from equation 2, we get :

$x+f^{\prime }(c)=0$

which is the singular solution in terms of x and y only.

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Example explanation

1. Solve $(y-px)(p-1)=p$ and obtain the singular solution.

We write it in Clairaut’s form.

$y=px+\frac{p}{p-1}$ Now we differentiate both sides w.r.t $x$.

$\frac{dy}{dx}=x.\frac{dp}{dx}+p+(\frac{1}{(p-1{)}^2}).\frac{dp}{dx}$ or $\frac{dy}{dx}=p+[x-\frac{1}{(1-p{)}^2}].\frac{dp}{dx}$ let $\frac{dy}{dx}=p$

Thus $\frac{dp}{dx}.[x-\frac{1}{(1-p{)}^2}]=0$ Thus from $\frac{dp}{dx}=0$, we get $p=c$.

Thus the required general solution is $y=cx+\frac{c}{c-1}$ And from $[x-\frac{1}{(1-p{)}^2}]=0$ we get $p=\frac{1+\sqrt{x}}{\sqrt{x}}$

We get singular solution as : $y=[\frac{1+\sqrt{x}}{\sqrt{x}}].x+\frac{1+\sqrt{x}}{\sqrt{x}}$ or $\boxed{y=2\sqrt{x}+x+1}$

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Second Order Differential Equations

A linear D.E of second order with constant coefficients is of the form: $\frac{d^2}{d{x}^2}+P_1.\frac{dy}{dx}+P_{2}y=x$

where $P_1$ and $P_2$ are constants and $x$ is either a constant or a function of $x$ itself.

We can solve this equation by first converting it to it’s auxiliary form using the D-operator method.

Using the symbol D for the differential operator $\frac{d}{dx}$ , the above equation can be written as:

$(D^2+P_{1}D+P_2)y=X$

Case 1: Solving homogenous part and finding C.F

Now if $X$ is equal to zero then the equation becomes homogenous. To solve this we need to find a complementary factor or C.F.

Rules for finding Complementary factor, C.F

1. Convert the given differential equation to it’s auxiliary form. For example : $D^2+D+1=0$

2. Solve the equation and find it’s two roots $m_1$ and $m_2$.

3. If the roots are different, the solution becomes: $y=c_1.e^{m_1.x}+c_2.e^{m_2.x}$ where $c_1$ and $c_2$ are arbitrary constants.

4. If the roots are same, i.e $m_1=m_2$, then the solution becomes: $y=[c_1+c_2.x]e^{mx}$

5. If the roots are complex then, the solution becomes: $y=e^{ax}[c_1.cos(bx)+c_2.sin(bx)]$ where $m=a+bi$ , $a=0$ , $b=1$.

   or $y=c_1.cos(bx)+c_2.sin(bx)$

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Case 2: Solving the non-homogenous part of the differential equation.

In the event of : $(D^2+P_{1}D+P_2)y=X$, $X$ is not a constant, there is a different way to find the solution to this equation.

Firstly we assume $y=e^{mx}$ to be a trial solution to this equation and find C.F for the homogenous part, and we write it down as :

$y=c_1.y_1(x)+c_2.y_2(x)$ Now for the non-homogenous part, we need to find something called a Particular-Integral or P.I.

The entire general solution will be

$y=C.F+P.I$

There are two ways to find the P.I .

Method 1: Using the inverse operator method.

Let the non-homogenous D.E be :

$(D^2+a_1.D+a_2)y=f(x)$

We do this by setting :

$P.I=\frac{1}{(D^2+a_1.D+a_2)}.f(x)$

Solving $(D^2+a_1.D+a_2)$ , we get either $(D-a{)}^n$ or $(D+a)$ or $(D-a)(D+b)$

Now a few important cases of $f(x)$ and formulae for those cases.

1. $\frac{1}{(D-a)}.X=e^{ax}.\int X.e^{-ax}dx$

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2. $\frac{1}{D}.X=\int Xdx$

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3. If you have :

   $\frac{1}{(D-m_1)(D+m_2)}.X$

   or

   $\frac{1}{f(D)}.X$

   , where $m_1$ and $m_2$ are roots of $f(D)$, then we can get the P.I as follows:

   $\frac{1}{f(D)}=\frac{A_1}{D-m_1}+\frac{A_2}{D-m_2}$

   which is solved as:

   $\frac{1}{f(D)}.X=A_1.e^{m_1.x}.\int X.e^{-m_1.x}+A_2.e^{m_2.x}.\int X.e^{-m_2.x}$

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Now to find the values of $A_1$ and $A_2$.

Let’s say we have a given expression

$\frac{1}{(D-1)(D+1)}.(x.e^{2x})$

where $m_1=1$ and $m_2=-1$.

We can write

$\frac{1}{(D-1)(D+1)}.(x.e^{2x})=[\frac{A}{D-1}+\frac{B}{D+1}].(x.e^{2x})$

or

$\frac{1}{(D-1)(D+1)}.(x.e^{2x})=\frac{A(D+1)+B(D-1)}{(D-1)(D+1)}.(x.e^{2x})$

Cancelling out $x.e^{2x}$ from both sides and multiplying both sides by $(D-1)(D+1)$,

We get

$1=\frac{A_1}{D-1}+\frac{A_2}{D+1}$

or
$1=(A+B)D+(A-B)$

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Now we compare both the coefficient of $D$ and constant terms to the other side.

Since there is no $D$ on the other side, $(A+B)=0$, and $1$ is on the other side, so

 $A-B = 1 => A = B$
 
 

From the two equations, we get $A=\frac{1}{2}$ , $B=-\frac{1}{2}$

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So we get our P.I as:

$\frac{1}{2}.e^x.\int x.e^{2x}.e^{-x}dx-\frac{1}{2}{e}^{-x}.\int x.e^{2x}.e^{x}dx$

which, when solved completely, gives us:

$\frac{e^{2x}}{(3x-4)}$

as the solution for this P.I.

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4. If you have :

   $\frac{1}{f(D)}.X$

   where $X=P(x)$, is a polynomial function of some degree $m$, we get the P.I as:

   $[f(D){]}^{-1}.P(x)$

   where we expand $D$ till $D^m$ and then solve $P(x)$.

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5. If you have:

   $\frac{1}{f(D)}.e^{ax}.V$

   , where $V$ is a function of a $x$. We can get the solution as:

   $P.I=e^{ax}.\frac{1}{f(D+a)}.V$

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Method 2: Using the method of variation of parameters.

This method is used when it’s difficult to apply the inverse operator method. However it is not a replacement for the inverse operator method. Both are valid, usage depends on the question.

So picking up from our C.F for equation: $(D^2+P_{1}D+P_2)y=X$

where we assumed the C.F to be: $y=c_1.y_1(x)+c_2.y_2(x)$ Now, let us assume our P.I as $y=u.y_1(x)+v.y_2(x)$, where $u$ and $v$ are unknown functions of $x$

To find the value of $u$ and $v$, we need to find the Wronskian Determinant, which is given by

$$\left|\begin{array}{c}{y}_1{y}_2\\ y_1^{\prime }{y}_2^{\prime }\end{array}\right|$$

and $u^{\prime }=\left|\begin{array}{c}0y_2\\ X{y}_2^{\prime }\end{array}\right|=\frac{-y_2.X}{W}$

and

$v^{\prime }=\left|\begin{array}{c}{y}_{1}0\\ y_1^{\prime }X\end{array}\right|=\frac{y_1.X}{W}$

or

$\boxed{u=\int \frac{-y_2.X}{W}}$ and

$\boxed{v=\int \frac{y_1.X}{W}}$

Put these values back into your P.I equation to get the full general solution to the D.E

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