MAT202 Test solution


Photo Credit: Mymathware

This is my very first tutorial on ordinary differential equation,

i am looking into some basic problems of O.D.E's and how to solve them.

This problems are obtained from

the continuous assessment of MAT202[Ordinary Differential Equations] course.

The problems include:

(i) Linear first order differential equation

(ii) How to evaluate homogeneous differential equations with

boundary conditions using Laplace transforms.

(iii) How to solve non-homogeneous linear differential equations using the method of

undetermined coefficients

(iv) And lastly, i also treated how to evaluate the

Laplace transforms of functions and how to evaluate the inverse Laplace transforms.  


(1) Solve $\frac{dy}{dx}+y=y^2$
Solution

This is a linear first order differential equation

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

$\frac{dy}{dx}=y(y-1)$

using variable separable form we get

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

you can verify the fact that

$(\frac{1}{y-1}-\frac{1}{y})dy=dx\Rightarrow\frac{dy}{dx}=y(y-1)$

we now integrate both sides of (1)

$\ln(y-1)-\ln(y)=x+C$ where $C$ is the constant of integration

by the laws of logarithm functions we get

$\ln(\frac{y-1}{y})=x+C$

$\frac{y-1}{y}=e^{x+C}$

$\frac{y-1}{y}=Ce^{x}$

now we need to obtain $y$

$y-1=yCe^x$

$y-yCe^x=1$

$y(1-Ce^x)=1$

$y=\frac{1}{1-Ce^x}$

and that's the required solution

(2) Solve the equation $y"+4y=0$, $y(0)=2$, $y'(0)=2$ by laplace transform method


Solution

This is a simple linear homogeneous differential equation,

but remember you do not have to solve it using the homogeneous method

because the questions says

"Use laplace transform method".

So we begin by taking the the laplace of the whole equation by:

$L\{y"\}+4L\{y\}=0$

and by the rules of taking laplace of a

differential equation we are going to have:

$=S^2L\{y\}-Sy(0)-y'(0)+4L\{y\}=0$

now we simply substitute the boundary conditions $y(0)=2$ and $y'(0)=2$ to get:

$S^2L\{y\}-2S-2+4L\{y\}=0$

rearrange the equation by collecting the like terms

$S^2L\{y\}+4L\{y\}-2S-2=0$

factor out $L\{y\}$

$L\{y\}[S^2+4]-2S-2=0$

$L\{y\}[S^2+4]=2S+2$

make $L\{y\}$ subject of the formula

$L\{y\}=\frac{2S+2}{S^2+4}$

Now take the lapalce inverse to obtain $y$

$y=L^{-1}\{\frac{2S+2}{S^2+4}\}$

Factor out $2$ from the numerator to get.

$y=2L^{-1}\{\frac{S+2}{S^2+4}\}$

separate the laplace function.

$=2L^{-1}\{\frac{S}{S^2+4}+\frac{2}{S^2+4}\}$

take the laplace inverse individually

$=2L^{-1}\{\frac{S}{S^2+4}\}+2L^{-1}\{\frac{2}{S^2+4}\}$

and by the properties of laplace transforms we get

$=2\cos{2t}+2\sin{2t}$



(3) Solve the equation using undetermined coefficient

$\frac{d^2}{dx^2y}-4\frac{dy}{dx}+3y=x^2$


Solution

This is a linear non-homogeneous differential equation,

and we can simply rewrite the equation as

$y"-4y'+3y=x^2$........(i) to get equation one

To solve this equation using undetermined coefficient,

there are three things involved namely

(i) Solving the homogeneous solution $Y_h$

(ii) Solving the particular solution $Y_p$


(iii) And the general Solution $Y=Y_h+Y_p$

Now we begin with the homogeneous solution

Let $y"-4y'+3y=0$

The characteristic equation of this differential equation is

$m^2-4m+3=0$

$(m-1)(m-1)=0$

$m=1$ and $m=3$

Since the characteristic equation has a real but

different root  then the homogeneous solution is:

$Y_h=C_{1}e^x+C_2e^{3x}$........(ii)

Now we proceed to the particular solution

Let $g(x)=x^2$, here we see that the polynomial at the

right hand side is a second degree

polynomial so we construct a second degree polynomial for $Y_p$

$Y_p=A_{2}x^2+A_{1}x+A_0$...........(iii)

we differentiate twice because the given D.E in the question is

a second order differential equation

$Y'_p=2A_2{x}+A_1$.......(iv)

$Y"_{p}=2A_2$.........(v)

substitute $Y_p$,$Y'_{p},Y"_{p}$ into (i)

$2A_2-4(2A_2x+A_1)+3(A_2x^2+A_1x+A_0)=x^2$

Now we equate the coefficient of the L.H.S to that of the R.H.S

$2A_2=1$............(vi)

$-8A_2-4A_1=0$......(vii)

$3A_2+3A_1+3A_0=0$..........(viii)

from (vi) we get $A_2=\frac{1}{2}$

substitute $A_2=\frac{1}{2}$ into (v)

$\Rightarrow{-8}(\frac{1}{2})-4A_1=0$

$=4A_1=4$

$=A_1=-1$

substitute $A_1$ and $A_2$ into (vi)

$\Rightarrow{3}(\frac{1}{2})+3(-1)+3A_0=0$

$\Rightarrow{A_0}=\frac{1}{3}$

Hence The values of the particular solution are $A_0=\frac{1}{3}$,

$A_1=-1$, and $A_2=\frac{1}{2}$

The particular solution is

$Y_p=\frac{1}{2}x^2-x+\frac{1}{3}$

And this leads us to a general solution

$Y=Y_h+Y_p$

$Y=Y_h=C_{1}e^x+C_2e^{3x}+\frac{1}{2}x^2-x+\frac{1}{3}$



(4)  (i) find the $L\{e^{at}\cos{bt}\}$


Solution


To evaluate this laplace transform without waste of time,

i will apply the shifting theorem

using the property of laplace transformation for $L\{e^{at}f(t)\}=F(s-a)$

if $L\{f(t)\}=F(s)$

and $L\{\cos{bt}\}=\frac{s}{s^2+b^2}$

then by the shifting theorem we have

$L\{e^{at}\cos{bt}\}=\frac{(s-a)}{(s-a)^2+b^2}$

(ii) Find $L^{-1}\{\frac{s}{s^2+6s+25}\}$
Solution

I will try to make the solution of this Laplace transform as easy as possible,

because many

people studying Laplace inverse find it difficult to evaluate,

so i will be simplifying the

transform to the easiest form, but in as much as i will be

simplifying it, you still need

basic knowledge of the Laplace transforms and algebraic simplifications.
$L^{-1}\{\frac{s}{s^2+6s+25}\}$
We can rewrite this as

=$L^{-1}\{\frac{s}{s^2+6s+9+16}\}$

$=L^{-1}\{\frac{s}{(s^2+6s+9)+4^2}\}$

$=L^{-1}\{\frac{s}{(s^2+3s+3s+9)+4^2}\}$

$=L^{-1}\{\frac{s}{(s^2+3s)+(3s+9)+4^2}\}$

$=L^{-1}\{\frac{s}{s(s+3)+3(s+3)+4^2}\}$

$=L^{-1}\{\frac{s}{(s+3)(s+3)+4^2}\}$

$=L^{-1}\{\frac{s}{(s+3)^2+4^2}\}$

this can be rewritten as

$=L^{-1}\{\frac{s+3-3}{(s+3)^2+4^2}\}$

we can further split this to get

$=L^{-1}\{\frac{s+3}{(s+3)^2+4^2}\}-L^{-1}\{{\frac{3}{(s+3)^2+4^2}}\}$

this can be rewritten as

$L^{-1}\{\frac{s+3}{(s+3)^2+4^2}\}-\frac{3}{4}L^{-1}\{\frac{4}{(s+3)^2+4^2}\}$

and by the properties of Laplace transform we have

$=L^{-1}\{e^{-3t}\cos{4t}\}-\frac{3}{4}L^{-1}\{e^{-3t}\sin{4t}\}$

$=\{e^{-3t}\cos{4t}\}-\frac{3}{4}\{e^{-3t}\sin{4t}\}$

Now we factor out $e^{-3t}$ to get

$=e^{-3t}\{\cos{4t}-\frac{3}{4}\sin{4t}\}$

And this brings us to the end of the solution.