Solving Differential Equations: Mathematical Solutions for Various Equations

In this document we will discuss about Solving Differential Equations and below are the summary points of this document:- The text includes mathematical equations and their solutions for various differential equations. It provides general solutions for different equations, considering integration factors and particular solutions. It concludes with specific solutions derived by substituting values and solving equations.

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 )