IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Different-ial eqns.........simple one

sankydreams

The solution of primitive integral eqn (x² + y²)dy = xydx, is y = y(x). If y(1) =1 and y(x₀) = e, then x₀is :

(a) sqroot of (2(e² - 1))

(b) sqroot of (2(e² + 1))

(c) e * root 3

(d) sqroot of ((e² + 1)/2)

Wats primitive integral eqn? Wats y = y(x)? Plz explain ur solution in detail for assured rates.....

ramyadiamond

Well, primitive integral eqn means the equation which gives the curve represented by that particular differential eqn. and y=y(x) means that y in the end is a function of x, and there shouln't be any term of y with x in such a way so that it cant be resolved.

Now, the given eqn is,

dy/dx = xy/x² +y²

this is a homogenous eqn, so write y=vx, and on differenciating w.r.t x, we get

dy/dx = v +x.dv/dx

substituting back in the eqn.,

v +x.dv/dx = vx²/ x² +v²x²

x.dv/dx = -v³ /1+v²

Now cross-multiply the terms, and integrate them,

1+v²/v³ dv = - 1/x dx

-1/2v² + ln v = -ln x

1/2v² = ln vx

substituting back v =y/x,

x²/2y² = ln y

c +x²/2 = y².ln y

Use the condition y(1)=1 to get c, and

c=-1/2

so the eqn becomes,

y².ln y =x²-1 /2

Now y(x₀) =e, so

e² =x₀² -1/2

hence x₀ = 2e²+1

sankydreams

ur answer is wrong ramya but i have still rated u as i got the correct answer by using ur method...........the answer is root3*e

Ask & Discuss Questions with Community & Experts