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.

Basic types of differential equations.

1. Linear Differential equation.

These can be of two types:

1. $$\boxed{{\displaystyle \frac{dy}{dx}+P(x).y=Q(x)}}$$

To solve this, we find an integrating factor(I.F).

And we find the solution in terms of y as:

2. $$\boxed{{\displaystyle \frac{dx}{dy}+P(y).x=Q(y)}}$$

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

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: $\boxed{{\displaystyle Mdx+Ndy=0}}$

This equation will be considered exact, if and only if

and the solution of the equation will be given by:

However there are times when :

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.

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

1. If equation is of the form $\boxed{{\displaystyle Mdx+Ndy=0}}$
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.

For the alternate form

1. If equation is of form $$\boxed{y(f(x,y)dx + xg(x,y)dy) = 0}$$
2. M = $y(f(x,y)$ , N = $xg(x,y)$.
3. If $$Mx-Ny \neq 0$$, 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.

3. Bernoulli's equation

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

Where $$y' = \frac{dy}{dx}$$ 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 :

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

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

Substitute back into the original equation, we get:

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

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:

or

Example explanation

Let's say we have the equation:

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

5. Integrate both the equations, we get:

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

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.

Example explanation

Let's say we have an equation.

So rewriting the equation we get :

So we differentiate both sides w.r.t $x$.

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

So we get the equation as :

which can be further reduced down to:

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:

Now we integrate both sides to get

which can be solved further down to get:

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

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

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

Example 2.

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

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

and then by solving further we get an equation:

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 $$2x{dp} - 2p dp = dx$$
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 :

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

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 $\frac{dx}{dy}=\frac{1}{p}$

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

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

or

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:

Clairaut's equation.

A differential equation of the form

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 $x$.

Let

or

Now we set both of the terms

and

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

which is the required general solution,

And from equation 2, we get :

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

Example explanation

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

We write it in Clairaut's form.

Now we differentiate both sides w.r.t $x$.

or

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 $$\boxed{y = 2\sqrt{x} + x + 1}$$

Second Order Differential Equations

A linear D.E of second order with constant coefficients is of the form:

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:

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 :

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

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

In the event of : $$(D^2 + P_1D + P_2)y = X$$, $X$ 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 :

Now for the non-homogenous part, we need to find something called a Particular-Integral or P.I.

The entire general solution will be

There are two ways to find the P.I .

Method 1: Using the inverse operator method.

Let the non-homogenous D.E be :

We do this by setting :

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

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

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:

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

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

Let's say we have a given expression

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

We can write

or

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

We get

or

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 $A=\frac{1}{2}$ , $B=-\frac{1}{2}$

So we get our P.I as:

which, when solved completely, gives us:

as the solution for this P.I.

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 $P(x)$.

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:

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

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: $$(D^2 + P_1D + P_2)y = X$$

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

and $$u' =
\begin{vmatrix}
0 \ y_2 \
X \ y'_2 \
\end{vmatrix}
= \frac{-y_2.X}{W}$$

and

or

and

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