Differential equation dy/dx=x*(2-y)^2

[w126]

Differential equation dy/dx = x * (2 - y)^2

I first separate the variables. dy/(2-y)^2=xdx
Next I integrated both sides getting (2-y)^-1=1/2x^2+c
I solve for y, I get an answer as 2-(1/(1/2)x^2+c)=y

The actual answer is 2-(1/x^2)+c

Does anyone know what I did wrong?
Thank you very much

 

[BigBeachBanana]

I first separate the variables. dy/(2-y)^2=xdx
Next I integrated both sides getting (2-y)^-1=1/2x^2+c
I solve for y, I get an answer as 2-(1/(1/2)x^2+c)=y

The actual answer is 2-(1/x^2)+c

Does anyone know what I did wrong?
Thank you very much

Your answer isn't necessarily wrong. You can always verify your solution by taking the derivative.

 

[Dr.Peterson]

Differential equation dy/dx = x * (2 - y)^2

I first separate the variables. dy/(2-y)^2=xdx
Next I integrated both sides getting (2-y)^-1=1/2x^2+c
I solve for y, I get an answer as 2-(1/(1/2)x^2+c)=y

The actual answer is 2-(1/x^2)+c

Does anyone know what I did wrong?
Thank you very much

Maybe their answer is wrong. (There is a difference between "the actual answer" and "the claimed answer"!) Or maybe you copied it incorrectly, just as you did your own, by misplacing parentheses. You presumably meant your answer to be

2-1/((1/2)x^2+c)=y​

Are you sure they didn't say y = 2-1/(x^2+c) ? That would at least be closer to the right answer.

Please show us an image of the problem and answer as given in the book.