Differential Equations - Formation of a Differential Equation and Problem Solving (Lecture-02)

Solved Examples

1. The solution of x3dydx+4x2tany=ex secy satisfying y ( 1)=0, is

(a) tany=(x2)exlogex

(b) siny=ex(x1)x4

(c) tany=(x1)exx3

(d) siny=ex(x1)x3

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Solution

The equation can be written as

cosydydx+4xsiny=exx3

put siny=zcosydydx=dzdx

dzdx+4zx=exx3

I.F =e4xdx4=x4

zx4=x4exx3dx+C

zx4=(x1)ex+C

x4siny=(x1)ex+C (put x=1,y=0)

siny=ex(x1)x4

Answer (b)

2. The solution of the equation

dydx+x(x+y)=x3(x+y)31 is

(a) 1x+y=x2+1+Cex

(b) 1(x+y)2=x2+1+Cex2

(c) 1(x+y)2=x+1+Cex

(d) 1x+y=x+1+Cex2

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Solution

 Put z=1(x+y)2 Now dzdx=2(x+y)3(1+dydx)

1(x+y)3(dydx+1)=x(x+y)2+x312dzdx=zx+x3dzdx2zx=2x3IF=e2xdx=ex2zex2=ex22x3dx+C=ettdt+C1(x+y)2ex2=tet+et+C1(x+y)2=x2+1+ex2

Answer (b)

3. The solution of differential equation

x=1+xydydx+x2y22!(dydx)2+x3y33!(dydx)3+ is

(a) y=logex+C

(b) y=(logex)2+C

(c) y=±logex+C

(d) xy=xy+C

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Solution

x=exydydxlogex=xydydxydy=1xlogexdx+Cy22=(logex)22+Cy=±(logex)2+Cy=±logex+C

Answer (c)

4. The function f(x) satisfying (f(x))2+4f(x)f(x)+(f(x))2=0 is given by

(a) ke(2+3)x

(b) ke(4±5)x

(c) ke(2±3)x

(d) None of these

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Solution

Solving we get

f(x)=4f(x)±16(f(x))24(f(x))22.1=2f(x)±3f(x)

f1(x)f(x)=2±3logef(x)=(2±3)x+Cf(x)=e(2±3)x+C=e(2±3)xeC=ke(2±3)x where k=eC

Answer (c)

5. A curve f(x) has normal at the point P(1,1) given by a(y1)+(x1)=0. If the slope of tangent at any point on the curve is proportional to the ordinate at that point, then the curve is

(a) y=eax1

(b) y1=eax

(c) y=ea(x1)

(d) None of these

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Solution

Normal at point P is ay +x=a+1

Slope of tangent at P=a=(dydx)(1,1)……………………..(1)

Now dydxαydydx=ky(dydx)(1,1)=k………………….(2)

From (1) & (2) k=a

dydx=aydyy=adx

logey=ax+C

Now C=a( as curve passes through (1,1))

logey=axa

y=ea(x1)

Answer (c)

6. If the length of portion of normal intercepted between the curve \& x-axis varies as the square of the ordinate, then the curve is

(a) kx+1k2x2= C. eky

(b) ky+k2y21=Cekx

(c) ky+k2x21=Cexy

(n) None of these

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Solution

Normal is Yy=dxdy(Xx)

Point on x-axis is (x+ydydx,0)

Now (ydydx+xx)2+(0y)2αy2

(ydydx)2+y2=ky2

dydx=k2y21

dyk2y21=dx

Integrate to get,

 loge |y+y21k2|=kx+Cky+k2y21=Cekx

Answer (b)

Exercise

1. Let y1(x)+y(x)g1(x)=g(x)g1(x),y(0)=0,xR, where f1(x) denotes ddxf(x) and g(x) is a given non-constant differentiable function on R with g(0)=g(2)=0. Then the value of y(2) is

(a) 2

(b) -1

(c) 0

(d) None of these

Show Answer Answer: c

2. The solution of the differential equation d3ydx38d2ydx2=0 satisfying y(0)=18,y1(0)=0 and y2(0)=1 is y1=e8x8x+7λ, then the numerical value of λ is

(a) 8

(b) 764

(c) 64

(d) 18

Show Answer Answer: c

3.* If the solution of the differential equation dydx+cosx(3cosy7sinx3)siny(3sinx7cosy+7)=0 is (sinx+cosy1)λ.(sinxcosy+1)μ=c, where c is arbitrary constant, then

(a) λ=5

(b) λ=2

(c) μ=2

(d) μ=5

Show Answer Answer: a,c

4.* The curve for which the area of the triangle formed by the x-axis, the tangent line and radius vector of the point of tangency is equal to a2 is

(a) x=Cy+a2y

(b) y=xCx2

(c) y=Cx+a2x

(d) x=Cya2y

Show Answer Answer: a,d

5.* The solution of (xdx+ydyxdyydx)=a2x2y2x2+y2 is

(a) x2+y2=asin((tan1y/x)+ constant )

(b) x2+y2=acos((tan1y/x)+ constant )

(c) x2+y2=a(tan(sin1y/x)+ constant )

(d) y=xtan( constant +sin11ax2+y2)

Show Answer Answer: a,d

6. If f(x)(f(x))2dx and f(1)=0, then f(x) is equal to

(a) (2(x1))1/4

(b) (5(x2))1/5

(c) (3(x1))1/3

(d) None of these

Show Answer Answer: c

7. Through any point ( x,y) of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. The curve, given that it divedes the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of

(a) circle

(b) parabola

(c) ellipse

(d) hyperbola

Show Answer Answer: b

8. Match the following

Column I Column II
(a) If the function y=e4x+2ex is a solution of the differential (p) 3
equation d3ydx313dydx=ky, then the value of k3 is
(b) Number of straight lines which satisfy the differential (q) 4
equation dydx+x(dydx)2y=0 is
(c) If real value of m for which the substitution, y=um will (r) 2
transform the differential equation, 2x4ydydx+y4=4x6
into a homogenious equation, then the value of 2 m is
(d) If the solution of differential equation x2d2ydx2+2xdydx=12y is (s) 1
y=Axm+Bxn, then |m+n| is
Show Answer Answer: aq;br;cp;ds

9. Read the following passage and answer the questions

If any differential equation be in the form f(f1(x,y)d(f1(x,y)+ϕ(f2(x,y))d(f2(x,y))+=0, then we can perform term by term integration. For e.g., sinxyd(xy)+(xy)d(xy)= cosxy+12(xy)2+C

(i) The solution of the differential equation (xy4+y)dxxdy=0 is

(a) x24+12(xy)2=C

(b) x44+13(xy)3=C

(c) x4412(xy)2=C

(d) x2412(xy)2=C

(ii) Solution of differential equation (2xcosy+y2cosx)dx+(2ysinxx2siny)dy=0 is

(a) x2cosy+y2sinx=C

(b) xcosyysinx=C

(c) x2cos2y+y2sin2y=C

(d) None of these

(iii) Solution of differential equation x21x2+y2(xdy+ydx)+y2(xdyydx)=0 is

(a) tan1xy12(yx)2=C

(b) tan1xy+12(yx)2=C

(c) tan1xy+13(yx)2=C

(d) None of these

Show Answer Answer: (i) b (ii) b (iii) c

10. Solution of the differentia equation (ex2+ey2) y dydx+ex2(xy2x)=0 is

(a) ex2(y21)+ey2=C

(b) ey2(x21)+ex2=C

(c) ey2(y21)+ex2=C

(d) ex2(y1)+ey2=C

Show Answer Answer: a

11. The solution of the differential equation (xcoty+logcosx)dy+(logsinyytanx)dx=0

(a) (sinx)y(cosy)x=C

(b) (siny)x(cosx)y=C

(c) (sinx)x(cosy)y=C

(d) None of these

Show Answer Answer: b

12. The solution of differential equation

x+ydydxyxdydx=xcos2(x2+y2)y3 is

(a) tan(x2+y2)=x2y2+C

(b) cot(x2+y2)=x2y2+C

(c) tan(x3+y3)=y2x2+C

(d) cot(x2+y2)=y2x2+C

Show Answer Answer: a


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