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The solution of the differential equation is :
dy/dx = y/x - (y2 /x2 + 1)½ …………………(1) the equation is homogenous,so y = Vx dy/dx = V + xdv/dx Put the value of dy/dx in eq(1) V + xdv/dx = V - (V2 + 1)½ = ∫-dv(v2 + 1)½ = ∫dx/x = log|v + (v2 + 1)½ | = -logx + logc = log|y/x + ((y/x)2 + 1)½ | = log c/x =|y/x + ((y/x)2 + 1)½ | = c/x
dy/dx = (x2 + 3y2 )/2xy ………….(1) Let y = vx dy/dx = v + xdv/dx Substitute the value of y and dy/dx in (1) v + x dv/dx = (1+3v2 )/2v x dv/dx = (1+3v2 )/2v - v x dv/dx = (1 + 3v2 - 2v2 )/2 x dv/dx = (1+ v2 )/2v 2v/(1+v2 ) dv = dx/x …………(2) Integrating both the sides ∫2v/(1+v2 ) dv = ∫dx/x Put t = 1 + v2 dt = 2vdv ∫dt/t = ∫dx/x =>log|t| = log|x| + log|c| =>log|t/x| = log|c| t/x = +- c (1+v2 )/x = +-c (1 + (y2 )/(x2 ))/x = +-c x2 + y2 = Cx3 ……….(3) y(1) = 3 1 + 9 = c(1)3 c = 10 From eq(3), we get x2 + y2 = 10x3
The first order, first degree differential equation y ’= f(x,y) is said to be homogeneous, if
Given x dy/dx = y(log y - log x+1) logy-logx=log(y/x) dy/dx=y/x(log(y/x) +1) substituting y=vx dy/dx=v+xdv/dx =>dv/(vlogv)=dx/x, integrating we get log(logv)=log(cx) log(y/x) = cx =>y/x = e(cx) =>y = xe(cx)
This is clearly a Homogenous differential equation, as RHS is expressed only in terms of y/x. To Solve this, Lety/x=t ⟹y=xt ⟹dy/dx = t+xdt/dx By (i) ⟹tant+t = t+xdt/dx ⟹tant=xdt/dx ⟹dx/x=dt/tant Integrating both sides, ⟹∫dx/x = ∫cot tdt ⟹logx = log(sint)+logC ⟹logx=log(C siny/x) ⟹x = Csin(y/x)
The solution of the differential equationis :
x -(xy)½ dy = ydx [x-(xy)½ ] - y = dx/dy dx/dy = x/y - [(xy)½ ]/y ……………….(1) Let V = x/y x = Vy dx/dy = V + ydv/dy ……..(2) V + ydv/dy = V - (V)½ ydv/dy = -(V)½ dv/(V)½ = -dy/y Integrating both the sides, we get (V(-½+1 ))/(-½+ 1) = -log y + c 2(V)½ = -log y + c 2(y/x)½ = -log y + c 2(x)½ = (y)½ (-log y + c) 2(x)½ = (y)1 /2log y + c(y)½ (x)½ = [(y)½ ]/2 log y + [c(y)½ ]/2
y/x cos y/x dx −(x/y sin y/x + cosy/x) dy = 0 ⇒dy/dx = (y/x cos y/x)/(x/y sin y/x + cos y/x) This is a homogeneous differential equation. Putting y = vx and dy/dx = v+xdv/dx, we have v+xdv/dx = (v2 cosv)/(sinv+vcosv) ⇒xdv/dx = (v2 cosv)/(sinv+vcosv) - v ⇒x dv/dx = (v2 cosv - v2 cosv - vsinv )/(sinv+vcosv) ⇒x dv/dx = - [v sinv/(sinv + vcosv)] ⇒∫[(sinv + vcosv)/v sinv]dv = ∫dx/x Integrating both the sides, we get ∫(cot v + 1/v)dv = - ln(x) + c ln(sin v) + ln(v) = -ln(x) + c ln(sin(y/x) + ln(y/x) + ln(x) = c ln(y/x sin y/x * x) + cy sin(y/x) = c
Find the differential equation of all the straight lines touching the circle x2 + y2 = r2 .
Let y = mx + c be the equation of all the straight lines touching the circle.
Given : The equation of the circle is x2 + y2 = r2 ---------->(1)
The tangent to the circle is c2 = r2 (1+m2 )
c = r √(1+m2 )
we know that y = mx + c---------->(2)
y = mx + r √(1+m2 ) ---------->(3)
y - mx = r √(1+m2 )
Differentiating wrt x we get dy/dx -m =0
dy/dx = m
Substituting this in equation (3)
y - (dy/dx . x) = r √(1+(dy/dx)2 )
Squaring on both sides, we get
y2 - (dy/dx . x)2 = [ r √(1+(dy/dx)2 )]2
[y - x(dy/dx)]2 = r2 (1+(dy/dx))2 is the required differential equation.
Answer: The differential equation of all the straight lines touching the circle x2 + y2 = r2 is [y - x(dy/dx)]2 = r2 (1+(dy/dx))2
The solution of differential equation x2 dy + y(x + y)dx = 0 when x = 1, y = 1 is:
x2 dy + (xy + y2 ) dx = 0 ⇒ x2 dy = - (xy + y2 )dx ⇒ .......(i) Let y = vx Differentiating w.r.t x we get Substituting the value of y and dy/dx in equation (1), we get: Integrating both sides, we get: ......(ii) Now, it is given that y = 1 at x = 1. Substituting D = 1/3 in equation (2), we get
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