x²y dx -(x³ + y³) dy = 0
Answers
solution
x²ydx-(x³+y³)dy=0
using homogeneous method
then we let y=vx
⇒dy/dx=(x³+y³)/x²y
dy/dx=(x³+ (vx)³)/x²(vx)
⇒ dy/dx=(x³+v³x³)/x³v
dy/dx=x³(1+v³)/x³v
dy/dx=(1+v³)/v
there for dy/dx=1+vdv/dx
⇒ 1+vdv/dx=(1+v³)/v
⇒vdv/dx=(1+v³)/v(-1)
dv/dx=(1/v+v²)-1
dv/dx=(1/v²+v-1/v)
⇒dv/(v²+v∧-1-v)=dx
⇒ (v³/3)+(inv)-(v²/2)=x
but v=y/x
⇒ (y/x)³/3 + (in(y/x)) - (y/x)²/2=x+c
Answer:
The solution of the given equation is
.
Step-by-step explanation:
Given the homogeneous differential equation,
Rewriting the equation,
Taking common from numerator and denominator of RHS
...(1)
Let
...(2)
Differentiating with respect to , we get
...(3)
Using equations (2) and (3) in equation (1),
Separating the variables,
Integrating on both sides,
Therefore, the required solution of the given equation is
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