Diferential equation (bernoulli)?

Question

how can i solve this diferencial equation, i think its bernoulli´s

t^2 dy/dt +y^2 =ty

book´s answer: e^(t/y) = ct

Answer 1

t²dy/dt+y^2=ty

dy/dt-y/t=-y²;          y/t=u;            u'=(ty'-y)/t²;          

t²u'=ty'-y;        y'=tu'+u;          tu'+u-u=-u²;       -du/u²=-dt/t

1/u=lnt+lnc=ln(ct)

ct=e^(1/u)=e^(t/y)

Answer 2

I don't understand why the constant of integration is multiplied by t (?)

But I got t=e^(t/y)+C

curiously enough.....  

Answer 3

I understand that Bernoulli equation is an equation of the form

dy/dt + yP(t) = f(t)y^n

What makes them special is that they can be reduced to a linear DE of the form

dy/dt + u*g(t) = h(t)

using the substitution

u = y^(1-n)

%%%%%%%%%%5

t^2 dy/dt +y^2 =ty

%%%%%%%%%%

First I divided both sides by t^2 to get

dy/dt + (t^-2) y^2 = (1/t)y

Then I rearranged the equation to get

dy/dt + (-1/t)y = -(t^-2) y^2

Now you have

u = y^(1 - 2) = y^-1

Differentiating both sides gives

du/dy = -1*y^-2 = -1/y^2

Solving for dy/du you have

dy/du = -y^2

By the Chain rule,

dy/dt = (dy/du)*(du/dt) = (-y^2)*(du/dt)

Substituting this into the equation I got

dy/dt - (1/t)y = -(t^-2) y^2

(-y^2)*(du/dt) - (1/t)y = -(t^-2) y^2

Multiplying both sides by -1/y^2 gives

du/dt + 1/yt = t^-2

Becuase u=y^-1 you have

du/dt + u(1/t) = t^-2

integrating factor = e^[∫dt/t] = e^ln|t| = t

Now you have

∫(d/dt)[ut] = ∫t^-1 dt

ut = ln|t| + ln|c| = ln|ct|

Because u = t/y

t/y = ln|ct|

e^(t/y) = ct