The differential equation (dy)/(dx)+P y=...

The differential equation `(dy)/(dx)+P y=Q y^n ,\ n >2` can be reduced to linear form y substituting `z=y^(n-1)` b. `z=y^n` c. `z=y^(n+1)` d. `z=y^(1-n)`

`z=y^(n-1)`

`z=y^(n)`

`z=y^(n+1)`

`z=y^(1-n)`

Text Solution

Verified by Experts

The correct Answer is:

Divide b `y^(n)`
`(1)/(y^(n))(dy)/(dx)+p.y^(1-n)=Q`
put`y^(1-n)=z`
`(1)/(y^(n))(dy)/(dx)=(1)/((1-n))(dz)/(dx)`
so equatio becomes
`(1)/((1-n))(dz)/(dx)+P.z=Q`

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