3okie.
q6)(2p+x)=(4y+x^{2})^{1/2} t=4y+x^{2}\Rightarrow dt/dx=4y'+2x
1/2dt/t^{1/2}=dx
A compilation of quesitons similar to the way bhargav had one :P (hence borrowed a part of the name..)
Keep solving and i will keep adding questions here...
To begin with first 5 are here...
Question 1) \left(\frac{dy}{dx}+1\right) x+\tan (x+y)=0
Question 2) f''(x)\left({1-f'(f(x))}\right) = f''(f(x)).(f'(x))^{2}
Question 3) xy' = y\ln(xy)
Question 4) y''=2x+(x^{2}-y')^{2}
Question 5) (1+x\sqrt{x^2+y^2})dx+(-1+\sqrt{x^2+y^2})ydy = 0
Question 6) x\frac{dy}{dx}+\left({\frac{dy}{dx}}\right)^2= y
Question 7) (x+y)(dx-dy)=dx+dy
Question 8) (2x^{2}+3y^{2}-7)xdx-(3x^{2}+2y^{2}-8)ydy = 0
Question 9) \frac{d^2y}{dx^2}(x^2+1) = 2x\frac{dy}{dx}
Question 10) \frac{dy}{dx}=\frac{yf'(x)-y^2}{f(x)}
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UP 0 DOWN 0 7 63
63 Answers
62=>lnx + c = -1/4 ln t -1/4t
I think this is not correct.. look at it a bit more closely...
3
1063. xy' = yln(xy)
let xy = t
y + xdy/dx = dt/dx
=> dt/dx -y = (t/x)ln(t)
=> dt/dx = (1+lnt)tx
=> dt(1+lnt)t = dxx
this is integrable i think
So, ln(xy) = kx -1
1WHICH QUESTIONS R STILL UNSOLVED???
1.let x+y=v
(dv/dx)x + tan v = 0
dv/tan v + dx/ log x =0
therefore vx=C
(x+y)x=c
3asish u made a simple mistake,i have also done the thing by substituting xy=t.
the final eqn will be
dt/dx-t/x=t/xln(t)
dt/(1+lnt)t=dx/x
106no my answer is correct,
ur answer can be simplified as
ln(1+lnxy) = lnkx
=> 1+ lnxy = kx
or ln(xy) = kx-1
106sankara,
in #23, answer 1
it is ln[sin(x+y)] + ln(x) = c
OR cosec(x+y) = kx
10618
dydx + yx = x2y3
==> -1y3dydx -1xy2 = -x2
Taking m=1/y2,
dm/dx = -2y3dydx
==> dm/dx - 2m/x = -2x2
this can be done by integrating factor method now.
solving it we get 1y2x2 + 2x =c
10616 can be done by integrating factor method (then one more by parts is needed)
11
xdy/dx -y = 3x2y+xy2
Substitute y=vx ==> dy/dx = v +xdv/dx
x(v + xdv/dx) - vx = 3vx3+3v2x3
=> dv/dx = 3vx + 3v2x = 3vx(1+v)
now use variable separation,
∫dvv(v+1) = ∫3xdx
10610
dydx = yf'(x)-y2f(x)
=> -1y2dydx = -f'(x)/y+1f(x)
(take 1/y = t so, dt/dx = -1y2dydx)
=> dt/dx = -tf'(x)f(x) + 1/f(x)
=> dtdx + tf'(x)f(x) = 1f(x)
This can be done by integrating factor method now.
ans: f(x)/y = x+c
116) dx/dt - 4x = cost
Integrating factor : e∫ - 4 dt = e - 4 t
now, x . e - 4 t = ∫ cost . e - 4 t dt
so, x . e - 4 t = 1/17 [ sint. e - 4 t - 4 cost . e - 4 t ] + c
so all questions r solved . waiting fr further questions.
421 ) Put y = xt dy / dx = t + x dt / dx
=> x dt/dx = - 4t / 1 + t
=>lnx + c = -1/4 ln t -1/4t
=> lnx + c = -1/4lny/x -y/4x c = -1/4 ( put x = y =1 )
Working further gives solution
18) put x2=m , so , dm = 2x dx ; y2=n so, dn = 2y . dx
now equation is : dm / dn = ( 3m + 2n - 8 ) / ( 2m + 3n - 7 )
putting m = t + \alpha , n = s + \beta ; dm = dt , dn = ds
ds / dt = [ 2( t + \alpha ) + 3 ( s + \beta ) - 7] / [3 ( t + \alpha ) + 2 ( + s ) - 8 )
now the equations involving \alpha , \beta , in the numerator and denominator are equated, to get \alpha = 2 and \beta = 1 respectively.
now equation becomes ds / dt = (2s + 3t ) / ( 3s + 2t )
now , s = ut is put. so , z + t dz / dt = ds / dt.
its replaced and an integrable equation is formed.
422 ) Pt of intersecn. of 2x - y + 1 = 0 & 2y - x - 1 = 0 is x = - 1/3 & y = 1/3
Put x=X - 1/3 & y = Y + 1/3
dY / dX = 2X -Y / 2Y - X
Put Y = tX
2t- 1 / 2(1 - t2) dt = dx / x
=> -1/2 ln(1-t2) - 1/2 ln(i + t/1-t) = ln x + c
To be simplified further...
423 ) Put y= xt
t + xdt/dx =t2 /( t - 1)
xdt/dx = t / (t - 1)
=> t - lnt = lnx + c
=> t - c = ln xt
Hence , t = ln xt + c
116.
taking x-y = t
1-dt/dx=(t+3/t+11)
dt/dx = 8/(t+11)
(t+11)2=16x+C
or (x-y+11)2=16x+c
424 ) Put x - y = t
=> dy / dx = 1 - dt/dx
1 - dt/dx = (t +3 )2 / (t + 11)2
=> dx = (t2 + 22t + 121) / ( 16 t + 112) dt
3q21)x^{2}dy+y^{2}dx+xy(dy+3dy)
=x^{2}dy+2xydx+d(xy^{2})+xy(dx-dy)
=d(x^{2}y)+d(xy^{2})+xy(dx-dy)
425)Put y = t sinx
y' = sinx dt/dx + tcosx
=> dt/dx + tcosx = 1/(2t2sin3x) + tcosx
=>2t3/3 = integ[ cosec3x ] + c
426 ) Put y = x2 t
y ' = 2xt + x3 dt/dx
=> x dt/dx = 2t (1 - t2) / (1 + t2)
=> ln x + c = integ [ 2t (i + t2 )/ (1 - t 2) ] dt
4@Nishant Sir Plzzz check our solutions n correct if any mistakes!!!!!!!!
2926)
write in the form 4x3 dx/dy - x4/y = y..
Now substitue x4 = z
so dz/dy - z/y = y
..now it can be easily integrated...Integrating Factor 1/y
Final answer x4 = y2 + cy