Problem 05 | Substitution Suggested by the Equation

Problem 05
dy/dx=sin(x+y)
 

Solution 05
dy/dx=sin(x+y)

dy=sin(x+y) dx
 

Let
z=x+y

dz=dx+dy

dy=dzdx
 

dzdx=sinz dx

dz=sinz dx+dx

dz=(sinz+1) dx

dzsinz+1=dx

dz1+sinz1sinz1sinz=dx

(1sinz) dz1sin2z=dx

(1sinz) dzcos2z=dx

 dzcos2zsinz dzcos2z=dx

Problem 03 | Substitution Suggested by the Equation

Problem 03
dy/dx=(9x+4y+1)2
 

Solution 03
dy/dx=(9x+4y+1)2

dy=(9x+4y+1)2 dx
 

Let
z=9x+4y+1

dz=9 dx+4 dy

dy=14(dz9 dx)
 

14(dz9 dx)=z2 dx

dz9 dx=4z2 dx

dz=4z2 dx+9 dx

dz=(4z2+9) dx

dz4z2+9=dx

dz(2z)2+32=dx

122dz(2z)2+32=dx

Problem 02 | Substitution Suggested by the Equation

Problem 02
siny(x+siny) dx+2x2cosy dy=0
 

Solution 02
siny(x+siny) dx+2x2cosy dy=0
 

Let
z=siny

dz=cosy dy
 

Hence,
z(x+z) dx+2x2 dz=0       → homogeneous equation
 

Let
z=vx

dz=v dx+x dv
 

vx(x+vx) dx+2x2(v dx+x dv)=0

x2(v+v2) dx+2vx2 dx+2x3 dv=0

Problem 01 | Substitution Suggested by the Equation

Problem 01
(3x2y+1) dx+(3x2y+3) dy=0
 

Solution 01
(3x2y+1) dx+(3x2y+3) dy=0
 

Let
z=3x2y

dz=3 dx2 dy

dy=12(3 dxdz)
 

Thus,
(z+1) dx+(z+3)[12(3 dxdz)]=0

(z+1) dx+32(z+3) dx12(z+3) dz=0

[(z+1)+(32z+92)] dx12(z+3) dz=0

(52z+112) dx12(z+3) dz=0

Substitution Suggested by the Equation | Bernoulli's Equation

Substitution Suggested by the Equation
Example 1

(2xy+1) dx3(2xy) dy=0
 

The quantity (2x - y) appears twice in the equation. Let
z=2xy

dz=2 dxdy

dy=2 dxdz
 

Substitute,
(z+1) dx3z(2 dxdz)=0

then continue solving.