reduce the D.E. dy/dx = cos(x+y) to the variable separable form and hence solve
Answers
Answered by
0
Answer:
i) 1+
dx
dy
=coses(x+y)
let x+y=t
1+
dx
dy
=
dx
dt
∴
dx
dt
= cosect
∫dx=∫sintdt
x=−cost+c
x=−cos(x+y)+c
(ii) (x−y)
2
dx
dy
=a
2
let (x−y)=t
∴1−
dx
dy
=
dx
dt
So we can write
t
2
(1−
dx
dt
)=a
2
1−
dx
dt
=
t
2
a
2
1−
t
2
a
2
=
dx
dt
∫dx=∫
1−
t
2
a
2
1
−dt
∫dx=∫
t
2
−a
2
t
2
dt
x=∫
t
2
−a
2
(t
2
−a
2
)+a
2
dt
x=∫dt+a
2
∫
t
2
−a
1
1
dt
x=t+a
2
2a
1
log∣
t+a
t−a
∣+c
x=t+
2
a
log∣
t+a
t−a
∣+c
x=(x−y)+
2
a
log∣
x−y+a
x−y−a
∣+c
Answered by
0
Answer:
(i) 1+dydx = coses (x+y) let x + y = t 1 + dydx = dtdx dtdx = ...
Step-by-step explanation:
Substitute u = x - y , we get du/d = 1 - dy/dx . The given...
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