Solving Differential Equations: Mathematical Solutions for Various 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 −x4 +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 −x4 + 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 usex=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