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151.

Form differential equation for `Ax^(2)-By^(2)=0` A)`xyy_(2)+x(y_(1))^(2)-yy_(1)=0` B)`y_(1)=y/x` C)`y_(1)=x/y` D)`y_(1)=xy`A. `xyy_(2)+x(y_(1))^(2)-yy_(1)=0`B. `y_(1)=y/x`C. `y_(1)=x/y`D. `y_(1)=xy`

Answer» Correct Answer - B
Discussion
152.

D.E. `yy_(1)+x=a`, where a is a constant, represents a family ofA. circles centred on X-axisB. circles centred on Y-axisC. parabolasD. ellipses

Answer» Correct Answer - A
Discussion
153.

`y^(2)=a(b-x)(b+x)` is a solution of the D.E. A)`y_(2)+x(y_(1))^(2)+yy_(1)=0` B)`xyy_(2)+x(y_(1))^(2)-yy_(1)=0` C)`xyy_(2)-x(y_(1))^(2)+yy_(1)=0` D)`yy_(1),y_(2)=x^(2)`A. `y_(2)+x(y_(1))^(2)+yy_(1)=0`B. `xyy_(2)+x(y_(1))^(2)-yy_(1)=0`C. `xyy_(2)-x(y_(1))^(2)+yy_(1)=0`D. `yy_(1),y_(2)=x^(2)`

Answer» Correct Answer - B
Discussion
154.

The order of the differential equation whosegeneral solution is given by `y=(C_1+C_2)cos(x+C_3)-C_4e^(x+4_5),`where `C_1,C_2,C_3,C_4,C_5`, are arbitrary constants, is(a)5(b) 4 (c)3 (d) 2A. (a) 5B. (b) 4C. (c) 3D. (d) 2

Answer» Correct Answer - (c)
Given, `y=(c_(1)+c_(2))cos (x+c_(3))-c_(4)e^(x+c_(5)) …. (i)`
` rArr y=(c_(1)+c_(2))cos (x+c_(3))-c_(4)e^(x)cdot e^(c_(5)) `
Now, let `c_(1)+c_(2)=A,c_(3)=B,c_(4)e^(c_(5))=c`
`rArr y=A cos (x+B)-ce^(x) …(ii)`
On differentiating w.r.t.x, we get
`dy/dx=-A sin (A+B)-ce^(x) …(iii)`
Again, on differentiating w.r.t.x, we get
`(d^(2)y)/dx^(2)=-Acos (x+B)-ce^(x) … (iv)`
`rArr (d^(2)y)/dx^(2)=-y-2ce^(x) … (v)`
`rArr (d^(2)y)/dx^(2)+y=-2ce^(x)`
Again, on differentiating w.r.t.x, we get
` (d^(3)y)/dx^(2)+dy/dx=-2ce^(x) ..(vi)`
`rArr (d^(3)y)/dx^(2)+dy/dx=(d^(2)y)/dx^(2)+y [from Eq. (v)]`
which is a differential equation of order 3.
Discussion
155.

The order of the differential equation, whose general solution is `y = C_1 e^x + C_2 e^(2x) + C_3 e^(3x) + C_4 e^(x-c_5)` , where `C_1, C_2, C_3, C_4, C_5` are arbitrary constants, isA. 2B. 3C. 4D. 5

Answer» Correct Answer - B
Discussion
156.

The order of the differential equation whosegeneral solution is given by `y=(C_1+C_2)cos(x+C_3)-C_4e^(x+4_5),`where `C_1,C_2,C_3,C_4,C_5`, are arbitrary constants, is(a)5(b) 4 (c)3 (d) 2A. 5B. 4C. 3D. 2

Answer» Correct Answer - C
Discussion
157.

Integrating factor of differential equation `cosx(dy)/(dx)+ysinx=1`is(a)`( b ) (c)cosx (d)`(e)(b) `( f ) (g)tanx (h)`(i)(c)`( d ) (e)secx (f)`(g)(d) `( h ) (i)sinx (j)`(k)A. `sin x`B. `sec x`C. `tan x`D. `cos x`

Answer» Correct Answer - B
We have,
`(dy)/(dx)+(tanx)y=secx`
`therefore" I.F. "=e^(int tan x dx)=e^(log secx)=sec x`
Discussion
158.

Integrating factor of differential equation `cosx(dy)/(dx)+ysinx=1`is(a)`( b ) (c)cosx (d)`(e)(b) `( f ) (g)tanx (h)`(i)(c)`( d ) (e)secx (f)`(g)(d) `( h ) (i)sinx (j)`(k)A. `cosx`B. `tanx`C. `secx`D. `sinx`

Answer» Correct Answer - C
`cosx(dy)/(dx)+ysinx=1`
or `(dy)/(dx)+y(sinx)/(cosx)=secx`
`therefore intPdx=int(sinx)/(cosx)dx`
`=-logcosx`
`=log secx`
`therefore` I.F. `=e^(intPdx)=e^(log secx)=secx`
Discussion
159.

Find the general solution of the differential equations:`(dx)/(dy)+secx y=tanx(0lt=x

Answer» `(dy)/(dx)+y sec x=tanx`
Here, `P=secx` and `Q= tanx`
`:. I.F.=e^(intPdx)=e^(intsecx)`
`=e^(log(secx+tanx))`
`=secx+tanx`
and general solution :
`y(secx+tanx)=inttanx(secx+tanx)dx+c`
`=int(secx+tanx+sec^(2)x-1)dx+c`
` impliesy(secx+tanx)=secx+tanx-1+c`
Discussion
160.

The solution of the differential equation `(xcoty + log cosx)dy +(logsiny-ytanx) dx=0` isA. `(sinx)^(y)(cosy)^(x)=c`B. `(siny)^(x)(cosx)^(y)=c`C. `(sinx)^(x)(cosy)^(y)=c`D. None of these

Answer» Correct Answer - B
`(xcoty+logcosx)dy+(log siny-ytanx)dx=0`
or `intd(xlogsiny)+intd(ylogcosx)=0`
or `xlogsiny+ylogcosx=logc`
or `(siny)^(x)(cosx)^(y)=c`
Discussion
161.

Solution of the differential equation `cosx dy= y(sinx -y)dx , 0 lt x lt pi/2` (A) `secx=(tanx+c)y` (B) `ysecx=tanx+c` (C) `ytanx=secx+c` (D) `tanx=(secx+c)y`

Answer» `dy/dx = (y(sinx - y))/cosx = y tan x - y^2 secx`
`1/y^2 dy/dx = 1/y tan x - secx`
let `1/y = t`
`-1/y^2 dy/dt = dt/dx`
now, `-dt/dx = t tan x - secx`
`- dt/dx - t tan x = -sec x`
`dt/dx + t tan x = sec x `
`IF = e^(int tan x dx)= e^(ln |secx|)= sec x`
`t(IF) = int (IF) sec x `
`t * sec x = int(sec^2 x)`
`t sec x = int sec^2 x = tan x+ c`
`1/y sec x = tan x + c`
`secx = y tanx + cy`
option 1 is correct
Discussion
162.

STATEMENT-1 : If the length of subtangent and subnormal at point (x, y) on y = f(x) are 9 and 4 then x is equal to + 6. and STATEMENT-2 : Product of subtangent and subnormal is square of the ordinate of the point.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True

Answer» Correct Answer - D
Discussion
163.

The solution of differential equations

Answer» Correct Answer - A(s), B(q), C(p), D(r)
Discussion
164.

Match the following

Answer» Correct Answer - A(p, q, t), B(r), C(q, t), D(q, s)
Discussion
165.

Solve `[xsin^2(y/x)-y]dx+x dy=0; y=pi/4`when `x=1.`

Answer» Correct Answer - `cot(y/x)=log_(e)|ex|`
`[xsin^(2)(y/x)-y]dx+xdy=0`
or `(dy)/(dx)=v+(xdv)/(dx)`
Therefore, given equation reduces to
`v+x(dv)/(dx)=v-sin^(2)v`
or `x(dv)/(dx) = -sin^(2)v`
or `"cosec "^(2)vdv=-(dx)/(x)`
Integrating both sides, we get
`-cotv=-log|x|-logC`
or `cot(y/x)=log|Cx|`...............(2)
Now, `y=pi/4` at `x=1`.
`therefore cot(pi/4) = log|C|`
or `1=log C`
or C=e
Substituting C=e in equation (2), we get
`cot(y/x)=log|ex|`
This is the required solution of the given differential equation
Discussion
166.

Match the following

Answer» Correct Answer - A(q), B(p), C(s), D(s)
Discussion
167.

If `xsin((y)/(x))dy = (y sin((y)/(x))-x)dx`and y(1) `= (pi)/(2)` then the value of `cos((y)/(e^(7)))` is __________.

Answer» Correct Answer - 7
Discussion
168.

` (x-y^(2)x)dx-(y-x^(2)y)dy=0`A. `1+y^(2)=c(1=x^(2))`B. `1-y^(2)=c(1+x^(2))`C. `1+y^(2)=c(1-x^(2))`D. `1-y^(2)=c(1-x^(2))`

Answer» Correct Answer - D
Discussion
169.

`(y^(2))/(x^(2))(dx)/(dy)=(y+1)/(x+1)`A. `y-x-log((x)/(y))=c`B. `(1)/(y)-(1)/(x)+log((x)/(y))=c`C. `y-x-logy+logx=c`D. `(y+y^(-1))-(x+x^(-1))=c`

Answer» Correct Answer - B
Discussion
170.

`(dy)/(dx)+(x y+y)/(x y+x)=0`A. `x+y+log((x)/(y))=c`B. `x-y+log((x)/(y))=c`C. `y-x+log((y)/(x))=c`D. `xlog(y+1)=ylog(x+1)+c`

Answer» Correct Answer - C
Discussion
171.

`(x^(2)-1)/(y-1)=xy(dy)/(dx)`A. `3x^(2)-6logx=2y^(3)-3y^(2)+c`B. `3x^(2)+6logx=3y^(3)-2y^(2)+c`C. `3x^(3)+6logx=y^(3)-y^(2)+c`D. `3x^(3)+6logx=2y^(3)-3y^(2)+c`

Answer» Correct Answer - A
Discussion
172.

`y^(3)dx-4dy=x^(2)dy`A. `tan^(-1)((x)/(2))=(1)/(y^(2))+c`B. `2tan^(-1)x=y^(-2)+c`C. `2tan^(-1)y=x^(-2)+c`D. `y^(4)-16y=x^(3)+c`

Answer» Correct Answer - A
Discussion
173.

Given two curves y = f(x) passing through (0, 1) and `y = int_(-oo)^(x) f(t) dt` passing through (0, 1/n). The tangents drawn to both the curves at the points with equal abscissas intersect on the x-axis, the curve is given byA. `y = e^(nx)`B. y = nxC. y = nlnxD. `y = nx^(2)`

Answer» Correct Answer - `f(x) = e^(nx)`
Discussion
174.

Solve`(d^2y)/(dx^2)=((dy)/(dx))^2`

Answer» `(d^(2)y)/(dx^(2))=((dy)/(dx))^(2)`
or `d/(dx)(dy)/(dx) = ((dy)/(dx))^(2)`
or `(d(dy)/(dx))/((dy)/(dx))=(dy)/(dx)dx`
or `int ((d(dy)/(dx)))/((dy)/(dx))=intdy`
`therefore log(dy)/(dx)=logy+logC_(1)`
`therefore (dy)/(dx)=C_(1)y`
`therefore (dy)/(y)=C_(1)dx`
`therefore int(dy)/(y) = C_(1)intdx`
`therefore logy=C_(1)x+k`
`therefore y=e^(C_(1)x+k)=C_(2)e^(C_(1)x)`
Discussion
175.

Solve the differential equation `x y(dx)/(y dx)=(1+y^2)/(1+x^2)(1+x+x^2)`

Answer» We have `(ydy)/(1+y^(2))=(1+x+x^(2))/(x(1+x^(2))`dx
Integrating both sides,
`int(ydy)/(1+y^(2))=int(1/x+1/(1+x^(2)))dx`
`therefore 1/2log_(e)(1+y^(2))=log_(e)x+tan^(-1)x+log_(e)C`
`log_(e)sqrt(1+y^(2))/(cx)=tan^(-1)x`
`rArr sqrt(1+y^(2))=cxe^(tan^(-1)x)`
Discussion
176.

Solve `(dy)/(dx) = (x+2y - 3)/(2x + y - 3)`

Answer» Here `a_(1) = 1, b_(1) = 2 " " :. (a_(1))/(b_(1)) = (1)/(2)`
`a_(2) = 2, b_(2) = 1 " " :. (a_(2))/(b_(2)) = 2`
Now, `(a_(1))/(b_(1)) != (a_(2))/(b_(2))`
Thus case - I applies, we have setting,
`{:(x=u+h),(y=v+k):}}`
`(dv)/(du) = ((u+2v) + (h+2k-3))/((2u + v) + (2h + k-3))0`
Choose h, k such that
`{:(h+2k-3=0),(2h + k - 3 =0):}}`
The solution is h = 1, k = 1
`rArr (dv)/(du) = (u+2v)/(2u + v)`
Set `v = tu, rArr (dv)/(du) = t + (dt)/(du)`
Our equation reduces to
`t + u(dt)/(du) = (u+2tu)/(2u + tu) = (1+2t)/(2+t)`
`rArr u(dt)/(du) = (1+2t)/(2+t) - t = (1+2t-2t-t^(2))/(2+t) = (1-t^(2))/(2+t)`
`rArr (2+t)/(1-t^(2)) dt = (du)/(2)`
`rArr (2)/(1-t^(2))dt+(1)/(2).2(t)/(1-t^(2)) dt = (du)/(u)`
On integrating, we get
`ln|(1+t)/(1-t)| - (1)/(2)ln|1-t^(2)| = ln|u|+ ln|k|`, k being the constant of integration.
`rArr ln|((1+(v)/(u))/(1-(v)/(u)))|-(1)/(2)ln(1-(v^(2))/(u^(2)))| = ln(uk)|`
`rArr |((u+v)/(u-v))| - ln|(sqrt(u^(2)-v^(2)))/(u)| = ln|(uk)|`
`rArr ln|((u+v)/(u-v).(u)/(sqrt(u^(2)-v^(2))))| = ln|uk|`
`rArr (usqrt(u+v))/((u-v)^((3)/(2))) = +- uk`
`rArr k(u-v)^((3)/(2)) = +-sqrt(u+v)`
`rArr k(x-1-y+1)^((3)/(2)) = sqrt(x-1+y-1)`
`rArr k(x-y)^((3)/(2)) = sqrt(x+y-2)`
`:. k^(2)(x-y)^(3) = (x+y-2)`
`rArr (x-y)^(3) = lambda(x+y-2), lambda` being another constant.
Discussion
177.

Solve `(x^(2) + y^(2))(dy)/(dx) = 2xy`

Answer» `(dy)/(dx) = (2xy)/(x^(2) + y^(2))`
Set `y = vx`, so that `(dy)/(dx) = v + x(dv)/(dx)`
`rArr x(dv)/(dx) = (2v)/(1+v^(2)) - v = (2v - v - v^(3))/(1+ v^(2)) = (v - v^(3))/(1 + v^(2))`
`rArr (dx)/(x) = (1 + v^(2))/(v(1-v^(2))) dv`
`rArr (dx)/(x) = ((1)/(v) + (1)/(1-v) - (1)/(1+v))dv` (Using partial fractions)
Integrating
`ln|x| = ln|v| - ln |1-v| - ln|1+v| + ln|k|`,
`rArr |x| = |(v)/((1-v)(1+v))||k|, k gt 0`
`rArr |x| = |((y)/(x))/(1-(y^(2))/(x^(2)))k|`
`rArr |x| = |(xy)/(x^(2)-y^(2))k|`
`rArr = ky = +- (x^(2) - y^(2))` where k is the constant of integration.
Discussion
178.

Solve `(dy)/(dx) = sec (x+y)`

Answer» Setting x+y-t, we have no differentiation w.r.t. x
`1+(dy)/(dx) = (dt)/(dx)`
Our equation not reads
`(dt)/(dx) - 1 = sec t`
`rArr (dt)/(dx) = 1+ sec t`
`rArr (dt)/(1+ sec t) = dx`
`rArr (cos t dt)/(1+cos t) = dx`
`rArr ((1+ cos t)-1)/(1+ cos t) dt = dx`
`rArr dt - (dt)/(1+ cos t) = dx`
`rArr dt - (dt)/(2 cos^(2).(t)/(2)) = dx`
`rArr dt-(1)/(2)sec^(2).(t)/(2) dt = dx`
Integrating, we get,
`t- tan.(t)/(2) = x + k`
`rArr x+y - tan.(x+y)/(2) =x+k`
`:. y - tan.(x+y)/(2) = k`, k being the constant of integration.
Discussion
179.

Solve `(x^(2) + 2xy + y^(2) + 1)(dy)/(dx) = 2(x+y)`

Answer» Observe that `x^(2) + 2xy + y^(2) = (x+y)^(2)`
The equation becomes,
`{(x+y)^(2)+1}(dy)/(dx) = 2(x+y)`
Set x+ y = t, so that `1+(dy)/(dx) = (dt)/(dx)` (on differentiation w.r.t. x)
Now, `(t^(2)+1)((dt)/(dx) - 1) = 2t`
`rArr (dt)/(dx) = (2t)/(t^(2) + 1) +1 = (t^(2) + 2t +1)/(t^(2)+1) = (t^(2) + 2t + 1)/(t^(2) + 1) = ((t + 1)^(2))/(t^(2)+1)`
`rArr (t^(2) + 1)/((t + 1)^(2))dt = dx`
`rArr ((t+1)^(2)-2(t+1)+2)/((t+1)^(2)) dt = dx`
`rArr dt - (2)/(t+1)dt + 2(dt)/((1+t)^(2)) = dx`
Integrating, we obtain,
`t-2ln |t+1| - (2)/(t-1) = x + k`
`rArr x+y - 2ln|x+y+1| - (1)/(x+y+1) = x+k`
`:. y- 2ln|x+y+1| - (2)/(x+y+1) = k`, k being the constant of integration
Discussion
180.

Find the equation of a curve passing through (1,1) and whose slope of tangent at a point (x, y) is `-(x)/(y)`.

Answer» `because` The slope of the tangent to the curve at a point (x, y) is `(dy)/(dx)`
`(dy)/(dx) = -(x)/(y)`
`rArr xdx + ydy = 0`
Integrating both sides we get `(x^(2))/(2)+(y^(2))/(2) = c`
`because` The curve passes through (1,1)
`:. (1)/(2) + (1)/(2) = c`
`rArr c = 1`
`:.` The equation of the curve is `x^(2) + y^(2) = 2`
Discussion
181.

Solve the following differential equation:`sqrt(1+x^2+y^2+x^2 y^2) +x y(dy)/(dx)=0`

Answer» Note that `f(x,y) = sqrt(1+x^(2)+y^(2)+x^(2)y^(2)) = sqrt((1+x^(2))(1+y^(2)))`
`sqrt(1+x^(2)) . Sqrt(1+y^(2)) = g(x)h(y)`
Thus, f(x,y) can be written as a product of two functions, one of x alone and the othr of y alone.
`sqrt(1+x^(2)+y^(2)+x^(2)y^(2))+ xy(dy)/(dx) = 0`
`rArr sqrt(1+ x^(2))sqrt(1+y^(2)) + xy (dy)/(dx) = 0`
`rArr (sqrt(1+x^(2)))/(x) dx + (y)/(sqrt(1+y^(2)))dy = 0`
Integration yields
`int (1+x^(2))/(xsqrt(1+x^(2)))dx+(1)/(2)int (d(1+y^(2)))/(sqrt(1+y^(2))) = k`
` rArr int (1)/(xsqrt(1+ x^(2)))dx+ int (x)/(sqrt(1+x^(2))) dx + (1)/(2) xx 2sqrt(1+y^(2)) = k`
`rArr int (-(1)/(2)dt)/(((1)/(2))sqrt(1+((1)/(2))^(2)))+ sqrt(1+ x^(2)) + sqrt(1+y^(2)) = k` [setting x = 1/t in the first integral].
`rArr - int (dt)/(sqrt(1+ t^(2))) + sqrt(1+x^(2)) + sqrt(1+y^(2)) = k`
`rarr -ln|t + sqrt(1+t^(2))| + sqrt(1+ x^(2)) + sqrt(1+ y^(2)) = k`
`rArr -ln |(1)/(x)+sqrt(1+(1)/(x^(2)))| + sqrt(1+x^(2)) + sqrt(1+y^(2)) = k`
`rArr sqrt(1+ x^(2)) - ln|(x+sqrt(1+x^(2)))/(x)| + sqrt(1+ y^(2)) = k`
which is the general solution of the differential equation.
Discussion
182.

Find the equation of the curve passing through (1, 1) and the slope of the tangent to curve at a point (x, y) is equal to the twice the sum of the abscissa and the ordinate.

Answer» `because` The slope of the tangent at (x, y) is `(dy)/(dx)`
`:. (dy)/(dx) = 2(x+y)` ...(1)
Let x + y = t
`rArr 1 + (dy)/(dx) = (dt)/(dx)`
`:.` From equation (1)
`(dt)/(dx) - 1 = 2t`
`rArr (dt)/(2t + 1) = dx`
Integrating both sides, we get
`(1)/(2) ln (2t + 1) = x + ln c_(1)`
`rArr ln(2t + 1) = 2x + ln c`, where `ln c = 2ln c_(1)`
`rArr 2t + 1 = ce^(2x)`
or `2(x + y) + 1 = ce^(2x)`
`because` The curve passes through (1, 1)
`:. 2(1+1) + 1= ce^(2)`
`rArr c = 5e^(-2)`
`:.` The equation of the curve is `2(x+y) + 1 = 5e^(2(x-1))`
Discussion
183.

The solution of the equation `(2+x)dy-ydx=0` represents a curve passing through a fixed point P, then the area of equilateral triangle with P as one vertex and `x+y=0` as its one side, isA. `2sqrt3`B. `sqrt3`C. `(2)/(sqrt3)`D. `(4)/(sqrt3)`

Answer» Correct Answer - C
We have,
`(2+x)dy-ydx=0rArr(1)/(y)dx-(1)/(2+x)dx=0`
On integrating, we obtain
`logy-log(x+2)=logC rArr y=C(x+2)`
Clearly, it represents a curve passing through a fixed point `P(-2,0)`.
The altitude of the equilateral triangle having one vertex at `P(-2,0)` and opposite side `x+y=0` is
`P=|(-2+0)/(sqrt(1+1))|=sqrt2`
So, its area is `(p^(2))/(sqrt3)=(2)/(sqrt3)` sq. units
Discussion
184.

If `y=f(x)` is the solution of equation `ydx+dy=-e^(x)y^(2)`dy, f(0)=1 and area bounded by the curve `y=f(x), y=e^(x)` and x=1 is A, thenA. curve y=f(x) is passing through `(-2,e)`.B. Curve `y=f(x)` is passing through `(1,1//e)`C. curve `y=f(x)` is passing through `(1,1//3)`D. `A=e+2/sqrt(e )-3`

Answer» Correct Answer - A::D
`ydx+dy=-e^(x)y^(2)dy`
`rArr (e^(-x)ydx+e^(-x)dy)/(y^(2))=-dy`
`rArr d(e^(-x)/y)=dy`
`rArr e^(-x)=y^(2)+cy`
`therefore f(0)=1, therefore c=0`
`rArr e^(-x)=y^(2)`
`rArr y=e^(-x//2)`
`rArr A = int_(0)^(1)(e^(x)-e^(-x//2))dx=e+2/sqrt(e)-3`
Discussion
185.

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is `y = 2(e^(x) - x-1)`.

Answer» We have , `(dy)/(dx) = y+2x`
`rArr (dy)/(dx) - y = 2x`
`IF = e^(-int dx) = e^(-x)`
Multiplying the equation by IF and integrating
`y xx e^(-x) = int 2xe^(-x) dx + k`, k being the constant of integration.
`= 2[x int e^(x) dx - int 1.(-e^(x))dx] + k`, using integration by parts
`= 2-xe^(-x) - 2e^(x) + k` ...(i)
As curve (i) passes through (0, 0)
0 = 0 - 2 + k
`:. k = 2`
Thus the curve is
`ye^(-x) = -2xe^(-x) - 2e^(-x) + 2`
`:. y = -2x - 2 + 2e^(x)`
`:. y = 2(e^(x) - x-1)`
Discussion
186.

In which of the following differential equationdegree is not defined?(a) `( b ) (c) (d)(( e ) (f) d^(( g )2( h ))( i ) y)/( j )(( k ) d (l) x^(( m )2( n ))( o ))( p ) (q)+3( r ) (s)(( t ) (u) (v)(( w ) dy)/( x )(( y ) dx)( z ) (aa) (bb))^(( c c )2( d d ))( e e )=xlog( f f )(( g g ) (hh) d^(( i i )2( j j ))( k k ) y)/( l l )(( m m ) d (nn) x^(( o o )2( p p ))( q q ))( r r ) (ss) (tt)`(uu)(vv)`( w w ) (xx) (yy)(( z z ) (aaa) d^(( b b b )2( c c c ))( d d d ) y)/( e e e )(( f f f ) d (ggg) x^(( h h h )2( i i i ))( j j j ))( k k k ) (lll)+( m m m ) (nnn)(( o o o ) (ppp) (qqq)(( r r r ) dy)/( s s s )(( t t t ) dx)( u u u ) (vvv) (www))^(( x x x )2( y y y ))( z z z )=xsin( a a a a )(( b b b b ) (cccc) d^(( d d d d )2( e e e e ))( f f f f ) y)/( g g g g )(( h h h h ) d (iiii) x^(( j j j j )2( k k k k ))( l l l l ))( m m m m ) (nnnn) (oooo)`(pppp)(qqqq)`( r r r r ) (ssss) x=sin(( t t t t ) (uuuu) (vvvv)(( w w w w ) dy)/( x x x x )(( y y y y ) dx)( z z z z ) (aaaaa)-2y (bbbbb)),|x|

Answer» Correct Answer - A::B
`x=sin((dy)/(dx)-2y) rArr (dy)/(dx) -2y=sin^(-1)x`
`x-2y=log(dy)/(dx) rArr (dy)/(dx) = e^(x-2y)`
Discussion
187.

The solution of `(x d d+y dy)/(x dy-y dx)=sqrt((1-x^2-y^2)/(x^2+y^2))`is(a) `( b ) (c)sqrt(( d ) (e) (f) x^(( g )2( h ))( i )+( j ) y^(( k )2( l ))( m ) (n))( o )=sin{( p ) (q)tan^(( r ) (s)-1( t ))( u )(( v ) (w) (x) y/( y ) x (z) (aa) (bb))+C}( c c )`(dd)(ee)`( f f ) (gg)sqrt(( h h ) (ii) (jj) x^(( k k )2( l l ))( m m )+( n n ) y^(( o o )2( p p ))( q q ) (rr))( s s )=cos{( t t ) (uu)tan^(( v v ) (ww)-1( x x ))( y y )(( z z ) (aaa) (bbb) y/( c c c ) x (ddd) (eee) (fff))+C}( g g g )`(hhh)(iii) `( j j j ) (kkk)sqrt(( l l l ) (mmm) (nnn) x^(( o o o )2( p p p ))( q q q )+( r r r ) y^(( s s s )2( t t t ))( u u u ) (vvv))( w w w )=(tan{s i (xxx) n^(( y y y ) (zzz)-1( a a a a ))( b b b b )(( c c c c ) (dddd) (eeee) y/( f f f f ) x (gggg) (hhhh) (iiii))+C}( j j j j )`(kkkk)(llll) `( m m m m ) (nnnn) y+xtan(( o o o o ) (pppp) c+( q q q q ) (rrrr)sin^(( s s s s ) (tttt)-1( u u u u ))( v v v v )sqrt(( w w w w ) (xxxx) (yyyy) x^(( z z z z )2( a a a a a ))( b b b b b )+( c c c c c ) y^(( d d d d d )2( e e e e e ))( f f f f f ) (ggggg))( h h h h h ) (iiiii))( j j j j j )`(kkkkk)A. `sqrt(x^(2)+y^(2))= sin{(tan^(-1)y//x)+C}`B. `sqrt(x^(2)+y^(2))=cos{(tan^(-1)y//x)+C}`C. `sqrt(x^(2)+y^(2))=(tan(sin^(-1)y//x)+C)`D. `y=xtan(c+sin^(-1)sqrt(x^(2)+y))`

Answer» Correct Answer - A::D
The differential equation can be rewritten as `(xdx+ydy)/sqrt(1-(x^(2)+y^(2)))=(xdy-ydx)/sqrt(x^(2)+y^(2))`
Since `dtan^(-1)(y//x)=(xdy-ydx)/(x^(2)+y^(2))`
and `d(x^(2)+y^(2))=2(xdx+ydy)`
We have `(1/2d(x^(2)+y^(2))/(sqrt(x^(2)+y^(2))sqrt(1-(x^(2)+y^(2)))))=(xdy-ydx)/(x^(2)+y^(2))=d{(tan^(-1)(y//x))}`
Put `x^(2)+y^(2)=r^(2)` in the LHS. Then
`(tdt)/(tsqrt(1-t^(2)))=d{tan^(-1)(y//x)}`
Integrating both sides, we get
`sin^(-1)t=tan^(-1)(y//x)+c`
i.e., `sin^(-1)sqrt((x^(2)+y^(2)))=tan^(-1)(y//x)+c`
i.e., `sin^(-1)sqrt((x^(2)+y^(2)))=tan^(-1)(y//x)+c`
Discussion
188.

Find the curves for which the length of normal is equal to the radius vector.A. circlesB. rectangular hyperbolaC. ellipsesD. straight lines

Answer» Correct Answer - A::B
We have length of the normal = radius vector
or `ysqrt(1+((dy)/(dx))^(2)) = sqrt((x^(2)+y^(2))`
or `y^(2)(1+((dy)/(dx))^(2)) = x^(2)+y^(2)`
or `x=+-y(dy)/(dx)`
i.e., `x=y(dy)/(dx)` or `x=-y(dy)/(dx)`
i.e., `x=y(dy)/(dx)` or `x=-y(dy)/(dx)`
i.e., `xdx-ydy=0` or `xdx+ydy=0`
Clearly, `x^(2)-y^(2)=c_(1)` represents a rectangular hyperbola and `x^(2)+y^(2)=c_(2)` represents circles.
Discussion
189.

Find the curves for which the length of normal is equal to the radius vector.

Answer» The length of normal ` |y| sqrt(y+((dy)/(dx))^(2))`
From the hypothesis of the problem
`|y| sqrt(1+((dy)/(dx))^(2))` = radius vector `= r = sqrt(x^(2) + y^(2))`
`rArr y^(2){1+ ((dy)/(dx))^(2)} = r^(2) = x^(2) + y^(2)`
`rArr y^(2) + y^(2)((dy)/(dx))^(2) = x^(2) + y^(2) " " rArr y^(2)((dy)/(dx))^(2) = x^(2)`
`rArr +- y(dy)/(dx) = x " " rArr +- ydy = xdx`
`rArr +- y^(2) = x^(2) - k^(2) " " :. x^(2) +- y^(2) = k^(2)`
This represents a circle or equilateral hyperbola according as we take + or - sign.
Discussion
190.

If `y=Asin mx+B cos mx`, then prove that`(d^2y)/(dx^2)+m^2y=0`

Answer» `y = Asinmx+Bcosmx`
`dy/dx = Amcosmx-Bmsinmx`
`(d^2y)/dx^2 = -Am^2sinmx-Bm^2cosmx`
`(d^2y)/dx^2 = -m^2(Asinmx+Bcosmx)`
`(d^2y)/dx^2 = -m^2(y)`
`(d^2y)/dx^2 + m^2(y) = 0`
Discussion
191.

Solve `(dy)/(dx) + x sin 2y = x^(3) cos^(2) y`

Answer» We have on dividing by `cos^(2)y`
`sec^(2)y(dy)/(dx) + ((2sin y cos y)/(cos^(2) y))x = x^(3)`
`rArr sec^(2)y(dy)/(dx) + (2 tan y) x = x^(3)`
Set `tan y = v`, so that `sec^(2) y (dy)/(dx) = (dv)/(dx)`
Our equation becomes,
`(dv)/(dx) + 2vx = x^(3)`
which is linear in v
`IF = e^(int 2x dx) = e^(x^(2))`
multiplying the equation by If and integrating
`ve^(x^(2)) = int e^(x^(2)). x^(3)dx + k`, k n being the constant of integration
`= int e^(x^(2)). x^(2).xdx + k` (set `t = x^(2)`)
`= (1)/(2) int e^(t). t dt + k`
`= (1)/(2)e^(t).(t-1) + k`
`rArr - tan y e^(x^(2)) = (1)/(2)e^(x^(2)) (x^(2)-1) + k`
`rArr 2 tan y e^(x^(2)) (x^(2)+1) + lambda, lambda` being another constant.
`:. 2 tan y = (x^(2) - 1) + lambda e^(-x^(2))`.
Discussion
192.

Find the differential equation of all lines in the `xy` - plane.

Answer» Equation of a line in `xy` plane,
`y = mx+c`
`:. dy/dx = m`
`=> (d^2y)/dx^2 = 0` ,which is the required equation.
Discussion
193.

Solve `(dy)/(dx) sqrt(1+x+y) =x+y-1`

Answer» Putting `sqrt(1+x+y)=v`, we have
`x+y-1=v^(2)-2`
or `1+(dy)/(dx)=2v(dv)/(dx)`
Then, the given transforms to
`(2v(dv)/(dx)-1)v=v^(2)-2`
or `(dv)/(dx)=(v^(2)+v-2)/(2v^(2))`
or `int(2v^(2))/(v^(2)+v-2)dv=intdx`
or `2int[1+1/(3(v-1))-4/(3(v+2))]dv=intdx`
or `2[v+1/3log|v-1|-4/3log|v+2|]=x+c`
where `v=sqrt(1+x+y)`
Discussion
194.

Solve : `(dy)/(dx) - (2y)/(x) = y^(4)`

Answer» On dividing the equation vy `y^(4)`, we have
`(1)/(y^(4)) (dy)/(dx) - ((2)/(y^(3)).(1)/(x)) = 1`
Set `(1)/(y^(3)) = v`, so that, `-(3)/(y^(4))(dy)/(dx) = (dv)/(dx)`
Our equation reads,
`-(1)/(3)(dv)/(dx) - (2v)/(x) = 1`
`rArr (dv)/(dx) + (6)/(x) v = -3`
which is a linear differential equation in v.
`IF = e^(6int (dx)/(x)) = e^(6 ln x) = x^(6)`
Multiplying the equation by IF and integrating, we get
`vx^(6) = -3int x^(6) dx + k`, k being a constant of integration.
`= -3 (x^(7))/(7) + k`
`rArr (1)/(y^(3)) x^(6) = -(3)/(7) x^(7) + k`
`rArr (3)/(7)x^(7) + (x^(6))/(y^(3)) = k`
`:. x^(6)(3x+(7)/(y^(3))) = lambda, lambda` being another constant.
Discussion
195.

Solve `(dy)/(dx)=(x+y)^2`

Answer» `(dy)/(dx) = (x+y)^(2)`………………(1)
Here direct variables separation is not possible.
So, we choose substitution method.
Let x+y=v.
`therefore 1+(dy)/(dx)=(dv)/(dx)`
So,equation (1), reduces to
`(dv)/(dx)=v^(2)+1`
or `int(dv)/(v^(2)+1)=intdx`
`therefore tan^(-1)v=x+c`
`rArr x+y=tan(x+c)`, which is required solution.
Discussion
196.

Solve `log(dy)/(dx)=4x-2y-2`, given that `y=1`when `x=1.`

Answer» `log_(e)(dy)/(dx)=4x-2y-2`
`rArr (dy)/(dx)=e^(4x-2y-2)`
`rArr e^(2y+2)dy=e^(4x)dx`
Integrating both sides,
`inte^(2y+2)dy=e^(4x)dx`
`rArr e^(2y_2)/(2)=e^(4x)/(4)+C`
Putting x=1 and y=1, we get
`e^(4)/2=e^(4)/4+C` or `c=e^(4)/4`
So, particular solution is `(e^(2y)+2)/(2) = (e^(4x)+e^(4))/(4)`
or `2(e^(2y+2))=e^(4x)+e^(4)`
Discussion
197.

Solve the following differential equation: `(1+y^2)dx=(tan^(-1)y-x)dy`

Answer» Rewrite the equation as,
`(dx)/(dy) = (tan^(-1)y)/(1+y^(2)) - (x)/(1+y^(2))`
`rArr (dx)/(dy) + (1)/(1+y^(2)) x = (tan^(-1)y)/(1+y^(2))`
This is linear in x, with from being `(dx)/(dy) + Rx = S`
IF `= e^(int Rdy) = e^(int(dy)/(1+y^(2))) = e^(tan^(-1)y)`
The solution is
`x. IF = int S. IF dy + k`, (k being the constant of integration)
`rArr x. e^(tan^(-1)y) = int e^(tan^(-1) y).(tan^(-1) y)/(1+ y^(2)) = dy` ...(i)
Put `tan^(-1)y = t rArr (dy)/(1+y^(2))= dt`
`:. int e^(tan^(-1)y)(tan^(-1)y)/(1+y^(2))dy = int t e^(1) dt = t int e^(t) dt - int e^(t) 1 dt`
`= te^(t) - e^(t) = (t-1)e^(t)`, (using Integration by parts)
From (i), `x.e^(tan^(-1)y) = (tan^(-1)y-1)e^(tan^(-1)y) + k`
`:. x = ke^(-tan^(-1)y) + tan^(-1) y- 1`
Discussion
198.

Solve the differential equation: `(1+x^2) dy/dx + y = tan^-1 x`

Answer» `(dy)/(dx)+y/(1+x^2)=tan^(-1)x/(1+x^2)`
P=`1/(1+x^2)` Q=`(tan^(-1)x)/(1+x^2)`
`y*IF=int(IF)*Q dx`
`y*e^(tan^(-1)x)=int(e^(tan^(-1)x))/(1+x^2) dx`
let`tan^(-1)x=t`
`y*e^t=(t-1)e^t+c`
`y=(tan^(-1)x-1)+ce^(-tan^(-1)x)`.
Discussion
199.

The differential equation for `y=Acos alphax +B sin alphax`, where A and B are arbitary constant isA. `(d^(2)y)/(dx^(2))-alpha .^(2)y=0`B. `(d^(2)y)/(dx^(2))+alpha .^(2)y=0`C. `(d^(2)y)/(dx^(2))+alphay=0`D. `(d^(2)y)/(dx^(2))-alphay=0`

Answer» Given, `y=A cos alpha+B sin alpha`
`Rightarrow (dy)/(dx)=-alpha A sin alpha x+alphaBcos x`
Again, differentiating both sides, w.r.t. x, we get
`Rightarrow(d^(2)y)/(dx^(2))=-Aalpha^(2)(Acos alphax+Bsin alphax)`
`Rightarrow(d^(2)y)/(dx^(2))=-alpha^(2)y`
`Rightarrow(d^(2)y)/(dx^(2))+alpha^(2)y=0`
Discussion
200.

If `y=e^(-x)(A cos x + B sin x)` then y is a situation ofA. `(d^(2)y)/(dx^(2))+2(dy)/(dx)=0`B. `(d^(2)y)/(dx^(2))-2(dy)/(dx)+2y=0`C. `(d^(2)y)/(dx^(2))+2(dy)/(dx)+2y=0`D. `(d^(2)y)/(dx^(2))+2y=0`

Answer» Given that, `y=e^(-x)(Acosx=Bsinx)`
On differentiating both sides w.r.t., x we get
`" "(dy)/(dx)=-e^(-x)(Acosx+Bcosx)+e^(-x)(-Asinx+Bcosx)`
`" "(dy)/(dx)=-y+e^(-x)(-Asinx+Bcosx)`
Again, differentiating both sides w.r.t., x, we get
`" "(d^(2)y)/(dx^(2))=(-dy)/(dx)+e^(-x)(-cosx-Bsinx)-e^(-x)(-Asin+Bcosx)`
`rArr" "(d^(2)y)/(dx^(2))=-(dy)/(dx)-y-[(dy)/(dx)+y]`
`rArr" "(d^(2)y)/(dx^(2))=-2(dy)/(dx)-2y`
`rArr" "(d^(2)y)/(dx^(2))+2(dy)/(dx)+2y=0`
Discussion