Question

If a curve y = f(x) satisfy the differential equation 2x? dy = (2xy + y*) dx and passes (1,2) then find f(1/2)

Answer 1

If a curve y = f(x) satisfy the differential equation 2x² dy = (2xy + y²)dx and passes (1, 2).

we have to find the value of f(1/2)

solution : here 2x² dy = (2xy + y²)dx

⇒dy/dx = (2xy + y²)/2x²

⇒dy/dx = y/x + y²/2x²

⇒1/(y²) dy/dx = 1/xy + 1/2x²

now let's put -1/y = p

differentiating w.r.t x , 1/y² dy/dx = dp/dx

⇒dp/dx = -p/x + 1/2x²

⇒dp/dx + p/x = 1/2x², it is the form of linear equation.

so, integrant factor, i.f = e^{∫(1/x) dx} = e^{lnx} = x

so the solution of linear equation is ...

p(i.f) = ∫(1/2x²) (i.f) dx

⇒-1/y × x = 1/2∫1/x² × x dx

⇒-x/y = 1/2 lnx + c

the curve passes through (1,2)

so, -1/2 = 1/2 ln1 + c

⇒c = -1/2

hence equation is -x/y = 1/2 lnx - 1/2

now at x = 1/2

-(1/2)/y(1/2) = 1/2 ln(1/2) - 1/2

⇒-1/y(1/2) = -ln2 - 1

⇒1/y(1/2) = ln2 + 1

⇒y(1/2) = 1/(ln2 + 1)

Therefore the value of f(1/2) = 1/(ln2 + 1)