Engineering Analysis 3: Solving General and Particular DEs

This assignment focuses on solving differential equations to find both general and particular solutions. It covers three main methods: direct integration, separation of variables, and the integrating factor method. Specific problems include finding general and particular solutions by direct integration and separation of variables, determining general and particular solutions of first-order differential equations using the integration factor method with given boundary conditions, and solving second-order homogeneous differential equations to determine general and particular solutions. The solutions involve applying integration techniques, substitution methods, and solving for constants using boundary conditions. Desklib offers a platform for students to access similar solved assignments and past papers for additional study resources.

1(a)
dy
dx −e3 x +4 x3=sin2 x
dy
dx = y '
y' =sin 2 x+ e3 x−4 x3
Now integrating both sides
∫ y' =¿ ∫ ( sin 2 x +e3 x−4 x3 ) dx ¿
y=−1
2 cos 2 x + e3 x
3 − 4
4 x4 +c
y=−1
2 c os 2 x + e3 x
3 −x4 +c
So general solution is as follow
y=−1
2 cos 2 x + e3 x
3 −x 4+ c
Now particular solution
Put x=0∧ y=0
0=−1
2 cos ( 0 ) + e0
3 −0+c
0=−1
2 + 1
3 +c
0=−1
6 +c
c= 1
6
So particular solution is as follow
y=−1
2 cos 2 x + e3 x
3 −x 4+ 1
6

1(b)
dy
dx = y +2
x +1
dy
y+ 2 = dx
x +1
∫ dy
y +2 =∫ dx
x+1
ln ( y +2 )=ln ( x+1 ) +lnc
ln ( y +2 )=ln c (x +1)
So general solution is as follow
ln ( y +2 )=ln c (x +1)
To find particular solution one has to use boundary conditions
ln ( y +2 )=ln c (x +1)
Now use x=0∧ y=0
ln ( 0+ 2 ) =ln c (0+1)
ln 2=ln c
So
c=2
ln ( y +2 )=ln 2(x+ 1)
Simplifying
y +2=2(x+1)
y=2 ( x +1 )−2
y=2 x +2−2
y=2 x
So particular solution is as follow

y=2 x
1(c)
( 3 y
9 x+ 3 )( dy
dx )= 5 y2+2
3 x2+ 2 x−6
( 3 y
5 y2+ 2 ) dy=
( 9 x +3
3 x2+ 2 x−6 ) dx
∫ ( 3 y
5 y2 +2 ) dy= ∫ ( 9 x +3
3 x2+2 x−6 ) dx
Now solving above two integrations separately
∫ ( 3 y
5 y2 +2 )dy this can be solved by using substitution method
Put u=5 y2 +2
du=10 y dy
y dy= du
10
∫ 3
10 ( du
u )
3
10 ln( u)
3
10 ln( 5 y2 +2)
Now solving second integration
∫ ( 9 x+ 3
3 x2 +2 x −6 ) dx=∫ ( 9 x
3 x2 +2 x −6 ) dx + ∫ ( 3
3 x2 +2 x−6 ) dx
After solving this we will get

∫ ( 9 x+ 3
3 x2 +2 x −6 ) dx= 3
2 ln (9 x2 +6 x−18)
So general solution is as follow
1
10 ln ( 5 y2 +2 )= 1
2 ln ( 9 x2 +6 x−18 ) + ln c
Now particular solution
1
10 ln ( 5+2 ) =1
2 ln ( 36+12−18 ) + ln c
1
10 ln ( 7 )=1
2 ln ( 30 ) +ln c
lnc= 1
10 ln 7− 1
2 ln30
So particular solution is as follow
1
10 ln ( 5 y2 +2 )= 1
2 ln ( 9 x2 +6 x−18 ) + 1
10 ln 7− 1
2 ln30
2(a)
dy
dx + y
x =x2+ 2
Integration factor

IF=e∫ 1dx
x
IF=elnx
IF=x
General solution is as follow
yx= ∫ ( x2+2 ) x dx+c
yx= ∫ ( x3+2 x ) dx+c
yx= x4
4 + 2 x2
2 +c
Now particular solution is as follow
1= 1
4 + 2
2 +c
c=−1
4
yx= x4
4 + x2− 1
4
y= ( x4 +4 x2−1
4 x )
2(b)
( x ) ( dy
dx )+ y=x cosx
Dividing by x on both sides
dy
dx + y
x =cos x

Integration factor
IF=e∫ 1dx
x
IF=elnx
IF=x
yx= ∫ cosx(x )d x+ c
yx=xsinx +cosx+c
y= ( xsinx+cosx+c
x )
Now particular solution is as follow
1= ( πsinπ +cosπ +c
π )
c=π +1
y= ( xsinx+cosx+ π +1
x )
(3)
2 ( d2 y
d x2 )−5 ( dy
dx ) −3 y=0
2 M2−5 M −3=0
So m1=3∧m2=−1
2
Roots are real so solution is as follow

y= A e3 x+ B e
−x
2
General solution is as follow
y= A e3 x+ B e
−x
2
Now to find particular solution
y= A e3 x+ B e
−x
2
0=A + B--------------------------------(1)
dy
dx =A e3 x 3−1
2 B e
− x
2
1=3 A−1
2 B---------------------------(2)
Multiplying equation (1) by ½
0= 1
2 A +1
2 B-----------------------------(3)
Now sloving equation (3) and (2)
A=2
7 ∧B=−2
7
So particular solution is as follow
y= 2
7 e3 x− 2
7 e
− x
2
y= 2
7 (e3 x−e
−x
2 )