Class 12th Math Differntial Equations Exercise 9.4 Ncert Solutions // Differential Equations Solutions

    Class 12^th Math Exercise 9.1 & 9.2 NcertSolutions // Differential Equations Solutions          

Class 12th Math Differntial Equations Exercise 9.4 Ncert Solutions  Differential Equations Solutions

          

     Differential Equations           

                                                              
     (Click Here For Exercise 9.1 & Exercise 9.2 )                                                       
                                                                 
                                    Exercise 9.4

For each of the differential equations in Exercises 1 to 10, find the general solution:

1.dydx=1−cosx1+cosxdydx=1−cos⁡x1+cos⁡x

Sol:    We have
dydx=1−cosx1+cosx=sin2x2+cos2x2−cos2x2+sin2x2sin2x2+cos2x2+cos2x2−sin2x2=sin2x2+sin2x2cos2x2+cos2x2=2sin2x22cos2x2=tan2x2dydx=(sec2x2−1)dydx=1−cos⁡x1+cos⁡x=sin2⁡x2+cos2⁡x2−cos2⁡x2+sin2⁡x2sin2⁡x2+cos2⁡x2+cos2⁡x2−sin2⁡x2=sin2⁡x2+sin2⁡x2cos2⁡x2+cos2⁡x2=2sin2⁡x22cos2⁡x2=tan2⁡x2dydx=(sec2⁡x2−1)

Separating the variables of the above eqaution we get
dy=(sec2x2−1)dxdy=(sec2⁡x2−1)dx

Now, integrating both sides of the above equation, we get
∫dy=∫(sec2x2−1)dxy=tan2x212−x+Cy=2tan2x2−x+C∫dy=∫(sec2⁡x2−1)dxy=tan2⁡x212−x+Cy=2tan2⁡x2−x+C
which is the general solution of the given differential equation


2.  dydx=√4−y2dydx=4−y2

Sol:    We have
dydx=√4−y2dydx=4−y2
Separating the variables of the above equation we get
dy√4−y2=dxdy4−y2=dx

Integrating both sides of the equation, we get

∫dy√4−y2=∫dx∫dy√22−y2=∫dxsin−1ya=x+Cya=sin(x+C)y=asin(x+C)∫dy4−y2=∫dx∫dy22−y2=∫dxsin−1⁡ya=x+Cya=sin(x+C)y=asin⁡(x+C)

which is the general solution of the given differential equation.

3. dydx+y=1(y≠1)dydx+y=1(y≠1)

Sol:    We have
dydx+y=1(y≠1)dydx=1−y−dyy−1=xdydx+y=1(y≠1)dydx=1−y−dyy−1=x

Separating the variables in the above equation, we get
−dyy−1=x−dyy−1=x

Integrating both sides of the equation we get
∫dyy−1=−∫xlog|y−1|=−x+Cy−1=e−x+Cy=1+e−x.eCy=1+Ae−x∫dyy−1=−∫xlog|y−1|=−x+Cy−1=e−x+Cy=1+e−x.eCy=1+Ae−x

4. sec2xtany dx+sec2ytanx dy=0sec2⁡xtan⁡y dx+sec2⁡ytan⁡x dy=0

Sol:    We have
sec2xtany dx+sec2ytanx dy=0sec2⁡xtan⁡y dx+sec2⁡ytan⁡x dy=0
Separating the variables of the above equation we get
sec2xtanx dx+sec2ytany dy=0sec2⁡xtan⁡x dx+sec2⁡ytan⁡y dy=0
Integrating both sides of the equation we get
∫sec2xtanx dx+∫sec2ytany dy=∫0∫d(tanx)tanx dx+∫d(tany)tany dy=∫0log|tanx|+log|tany|=logClog|tanxtany|=logCtanxtany=C∫sec2⁡xtan⁡x dx+∫sec2⁡ytan⁡y dy=∫0∫d(tan⁡x)tan⁡x dx+∫d(tan⁡y)tan⁡y dy=∫0log|tan⁡x|+log|tan⁡y|=log⁡Clog|tan⁡xtan⁡y|=log⁡Ctan⁡xtan⁡y=C
which is the general solution of the given differential equation


Class 12th Math Differntial Equations Exercise 9.4 Ncert Solutions  Differential Equations Solutions

(Click Here For Exercise 9.1 & Exercise 9.2 ) 

Top 20 Important 100% Common Questions And Answers For H.S Final 2020(Click Here)For English

5.(ex+e−x)dy−(ex−e−x)dx=0(ex+e−x)dy−(ex−e−x)dx=0

Sol:   We have
(ex+e−x)dy−(ex−e−x)dx=0(ex+e−x)dy−(ex−e−x)dx=0
Separating the variables of the above equation we get
dy=ex+e−xex−e−xdxdy=ex+e−xex−e−xdx

Integrating both sides of the above equation, we get
∫dy=∫ex+e−xex−e−xdx∫dy=∫d(ex−e−x)ex−e−xdxy=log|ex−e−x|+C∫dy=∫ex+e−xex−e−xdx∫dy=∫d(ex−e−x)ex−e−xdxy=log⁡|ex−e−x|+C
which is the general solution of the given differential equation.

6. dydx=(1+x2)(1+y2)dydx=(1+x2)(1+y2)

Sol:    We havedydx=(1+x2)(1+y2)dydx=(1+x2)(1+y2)
Separating the variables of the above equation we get
dy1+y2=(1+x2)dxdy1+y2=(1+x2)dx
Integrating both sides of the above equation we get
∫dy1+y2=∫(1+x2)dx∫dy1+y2=∫dx+∫x2dxtan−1y=x+x33+C∫dy1+y2=∫(1+x2)dx∫dy1+y2=∫dx+∫x2dxtan−1⁡y=x+x33+C
 which is the general solution of the given differential equation


7.  ylogy dx−x dy=0ylog⁡y dx−x dy=0

Sol:  We haveylogy dx−x dy=0ylog⁡y dx−x dy=0
Separating the variables of the above equation we get
dyylogy=dxxdyylog⁡y=dxx
Integrating both sides of the above equation,we get
∫dyylogy=∫dxx∫d(logy)logy=∫dxxlog|logy|=logx+logClog|logy|=logxClogy=xCy=exC∫dyylog⁡y=∫dxx∫d(log⁡y)log⁡y=∫dxxlog⁡|log⁡y|=log⁡x+log⁡Clog⁡|log⁡y|=log⁡xClog⁡y=xCy=exC
which is the genral solution of the given differential equation


8. x5dydx=−y5x5dydx=−y5

Sol: We havex5dydx=−y5x5dydx=−y5
Separating the variables of the above equation we get
dyy5+dxx5=0dyy5+dxx5=0
Integrating both sides of the above equation we get
∫dyy5+∫dxx5=Ay−5+1−5+1+x−5+1−5+1=Ay−4−4+y−4−4=Ay−4+x−4=−4Ay−4+x−4=C∫dyy5+∫dxx5=Ay−5+1−5+1+x−5+1−5+1=Ay−4−4+y−4−4=Ay−4+x−4=−4Ay−4+x−4=C
which is the required general solution of the given differential equation


9. dydx=sin−1xdydx=sin−1⁡x

Sol: We havedydx=sin−1xdydx=sin−1⁡x
Separating the variables in the above equation we have
 dy=sin−1x dxdy=sin−1⁡x dx
Integrating both sides of the above equation we get
∫dy=∫sin−1x dxy=sin−1x∫1dx−∫1√1−x2 dx (Integrating by parts)y=xsin−1x+∫−2x2√1−x2 dxy=xsin−1x+∫d(√1−x2)y=xsin−1x+√1−x2+C∫dy=∫sin−1⁡x dxy=sin−1⁡x∫1dx−∫11−x2 dx (Integrating by parts)y=xsin−1⁡x+∫−2x21−x2 dxy=xsin−1⁡x+∫d(1−x2)y=xsin−1⁡x+1−x2+C
which is the general solution of the given differential equation.

10 extany dx+(1−ex)sec2y dy=0extan⁡y dx+(1−ex)sec2⁡y dy=0

Sol: We have extany dx+(1−ex)sec2y dy=0extan⁡y dx+(1−ex)sec2⁡y dy=0
Separating the variables, the above equation can be written as
(ex1−ex)dx+(sec2ytany)dy=0(ex1−ex)dx+(sec2⁡ytan⁡y)dy=0
Integrating both sides of the above equation we get
∫(ex1−ex)dx+∫(sec2ytany)dy=∫0∫−(−ex1−ex)dx+∫(sec2ytany)dy=∫0∫(d(1−ex)1−ex)dx+∫(d(tany)tany)dy=∫0−log|1−ex|+log|tany|=logClog(tany1−ex)=logCtany=C(1−ex)∫(ex1−ex)dx+∫(sec2⁡ytan⁡y)dy=∫0∫−(−ex1−ex)dx+∫(sec2⁡ytan⁡y)dy=∫0∫(d(1−ex)1−ex)dx+∫(d(tan⁡y)tan⁡y)dy=∫0−log|1−ex|+log|tan⁡y|=log⁡Clog⁡(tan⁡y1−ex)=log⁡Ctan⁡y=C(1−ex)
which is the general solution of the given differential equation.


(Click Here For Exercise 9.1 & Exercise 9.2 ) 

11. x3+x2+x+1)dydx=2x2+x:y=1x3+x2+x+1)dydx=2x2+x:y=1 when x =0

Sol: We have(x3+x2+x+1)dydx=2x2+x(x3+x2+x+1)dydx=2x2+x
Separating the variables the above equation can be written as
dy=2x2+xx3+x2+x+1 dxdy=2x2+xx3+x2+x+1 dx
Integrating both sides of the equation we get
∫dy=∫2x2+xx3+x2+x+1 dxy=∫2x2+x(x2+1)(x+1) dxLet 2x2+x(x2+1)(x+1)=Ax+1+Bx+Cx2+12x2+x(x2+1)(x+1)=A(x2+1)+(Bx+C)(x+1)(x+1)(x2+1)2x2+x=A(x2+1)+(Bx+C)(x+1)2x2+x=Ax2+Bx2+Bx+Cx+A+CEquating the coefficients of corresponding terms we getA+B=2B+C=1A+C=0Solving the above three equations we getA=12,  B=32,  C=−12Substituting these values in eq(1) we gety=∫12(x+1) dx+∫3x−12(x2+1) dxy=∫12(x+1) dx+∫3x2(x2+1) dx−∫12(x2+1) dxy=∫12(x+1) dx+34∫2x(x2+1) dx−12∫1(x2+1) dxy=12log|x+1|+34log|x2+1|−12tan−1x+C(1)(1)∫dy=∫2x2+xx3+x2+x+1 dx(1)y=∫2x2+x(x2+1)(x+1) dxLet 2x2+x(x2+1)(x+1)=Ax+1+Bx+Cx2+12x2+x(x2+1)(x+1)=A(x2+1)+(Bx+C)(x+1)(x+1)(x2+1)2x2+x=A(x2+1)+(Bx+C)(x+1)2x2+x=Ax2+Bx2+Bx+Cx+A+CEquating the coefficients of corresponding terms we getA+B=2B+C=1A+C=0Solving the above three equations we getA=12,  B=32,  C=−12Substituting these values in eq(1) we gety=∫12(x+1) dx+∫3x−12(x2+1) dxy=∫12(x+1) dx+∫3x2(x2+1) dx−∫12(x2+1) dxy=∫12(x+1) dx+34∫2x(x2+1) dx−12∫1(x2+1) dx(1)y=12log|x+1|+34log|x2+1|−12tan−1⁡x+C

Given that y=1 when x=0. Substituting these values in above equation we get
1=12log|0+1|+34log|02+1|−12tan−10+CC=11=12log|0+1|+34log|02+1|−12tan−1⁡0+CC=1
Substituting the value of C in eq (1) we get
y=12log|x+1|+34log|x2+1|−12tan−1x+1y=14log|x+1|2+14log|x2+1|3−12tan−1x+1y=14log[(x+1)2(x2+1)3]−12tan−1x+1y=12log|x+1|+34log|x2+1|−12tan−1⁡x+1y=14log|x+1|2+14log|x2+1|3−12tan−1⁡x+1y=14log⁡[(x+1)2(x2+1)3]−12tan−1⁡x+1

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12. x(x2−1)dydx=1:y=0x(x2−1)dydx=1:y=0 when x =2

Sol:  We have
x(x2−1)dydx=1x(x2−1)dydx=1
Separating the variables the above equation can be written as
dy=dxx(x+1)(x−1)dy=dxx(x+1)(x−1)
Integrating both sides of the above equation, we get
∫dy=∫dxx(x+1)(x−1)∫dy=∫(−1x+12(x+1)+12(x−1) dx)y=−logx+12log+1)+12log−1)+logky=12log[k2(x+1)(x−1)x2](1)∫dy=∫dxx(x+1)(x−1)∫dy=∫(−1x+12(x+1)+12(x−1) dx)y=−log⁡x+12log+1)+12log−1)+log⁡k(1)y=12log⁡[k2(x+1)(x−1)x2]
Given that y=0 when x=2. Substituting these values in eq (1) we get
0=12log[k2(2+1)(2−1)22] 3k24=e0k2=430=12log⁡[k2(2+1)(2−1)22] 3k24=e0k2=43
Substituting this value of k in eq (1) we get
y=12log[4(x+1)(x−1)3x2]y=12log⁡[4(x+1)(x−1)3x2]

Class 12 Ncert Solution Maths Differential Equations

13. cos(dydx)=a(a∈R)cos⁡(dydx)=a(a∈R);y =1 when x =0

Sol: We have
cos(dydx)=acos⁡(dydx)=a
Separating variables the above equation can be written as
dydx=cos−1ady=cos−1a dx(1)dydx=cos−1⁡a(1)dy=cos−1⁡a dx
Integrating both sides of eq (1) we get
∫dy=∫cos−1a dxy=xcos−1a+C(2)(2)∫dy=∫cos−1⁡a dxy=xcos−1⁡a+C
Given that y=1 when x = 0. Putting these values in eq (2) we get
1=0×cos−1a+CC=11=0×cos−1⁡a+CC=1
Substituting the value of C in eq (2) we get
y=xcos−1a+1y−1=xcos−1ay−1x=cos−1acosy−1x=ay=xcos−1⁡a+1y−1=xcos−1⁡ay−1x=cos−1⁡acos⁡y−1x=a

14. dydx=ytanxdydx=ytan⁡x ;y=1 when x =0

Sol: We havedydx=ytanxdydx=ytan⁡x
Separating the variables the baove equation can be written as
dyy=tanx dxdyy=tan⁡x dx
Integrating both sides of the above equation we get
∫dyy=∫tanx dxlogy=log(secx)+logClogy=log(Csecx)y=Csecx(1)∫dyy=∫tan⁡x dxlog⁡y=log⁡(sec⁡x)+logClog⁡y=log⁡(Csec⁡x)(1)y=Csec⁡x
Given that y =1 when x =0. Substituting these values in eq(1) we get
1=Csec0C=11=Csec⁡0C=1
Substituting this value of C in (1) we get
y=secxy=sec⁡x

15. Find the equation of a curve passing through the point (0,0) and whose differential equation is y′=exsinxy′=exsin⁡x.

Sol: The given differential equation is
y′=exsinxy′=exsin⁡x
dydx=exsinxdydx=exsin⁡x
dy=exsinxdxdy=exsin⁡xdx

Integrating both sides of the equation, we get
∫dy=∫exsinxdxy=sinx ex−∫excosx dx (Integration by parts)y=sinx ex−[cosxex−∫(−sinx)ex dx]y=sinx ex−[cosxex+∫(sinx)ex dx]y=sinx ex−cosxex−∫(sinx)ex dxy=ex(sinx−cosx)−yy+y=ex(sinx−cosx)2y=ex(sinx−cosx)y=12ex(sinx−cosx)+C(1)∫dy=∫exsin⁡xdxy=sin⁡x ex−∫excos⁡x dx (Integration by parts)y=sin⁡x ex−[cos⁡xex−∫(−sin⁡x)ex dx]y=sin⁡x ex−[cos⁡xex+∫(sin⁡x)ex dx]y=sin⁡x ex−cos⁡xex−∫(sin⁡x)ex dxy=ex(sin⁡x−cos⁡x)−yy+y=ex(sin⁡x−cos⁡x)2y=ex(sin⁡x−cos⁡x)(1)y=12ex(sin⁡x−cos⁡x)+C
Substituting x=0 and y=0 in the above equation we get
0=12e0(sin0−cos0)+C0=121(−1)+CC=−120=12e0(sin⁡0−cos⁡0)+C0=121(−1)+CC=−12
Substituting the value of C in equation(1) we get
y=12ex(sinx−cosx)+122y=12ex(sinx−cosx)+12y−1=12ex(sinx−cosx)y=12ex(sin⁡x−cos⁡x)+122y=12ex(sin⁡x−cos⁡x)+12y−1=12ex(sin⁡x−cos⁡x)
 which is the required equation of a curve passing throught the point (0,0).

(Click Here For Exercise 9.1 & Exercise 9.2 ) 

Class 12th Math Differntial Equations Exercise 9.4 Ncert Solutions // Differential Equations Solutions

16. For the differential equation xydydx=(x+2)(y+2)xydydx=(x+2)(y+2), find the solution curve passing through the point (1,-1).

Sol: We havexydydx=(x+2)(y+2)xydydx=(x+2)(y+2)
Separating the variables the above equation can be written as
ydyy+2=x+2x dxydyy+2=x+2x dx
Integrating both sides of the above equation, we get
∫ydyy+2=∫x+2x dx∫y+2−2y+2 dy=∫x+2xdx∫1.dy−21y+2 dy=∫dx+21x dxy−2log|y+2|=x+2logx+C(1)∫ydyy+2=∫x+2x dx∫y+2−2y+2 dy=∫x+2xdx∫1.dy−21y+2 dy=∫dx+21x dx(1)y−2log⁡|y+2|=x+2log⁡x+C
The curve passes through the point (1, -1). Substituting  x =1 and y = -1 in eq (1) we get
−1−2log|−1+2|=1+2log1+CC=−2 since log 1 =0−1−2log⁡|−1+2|=1+2log⁡1+CC=−2 since log 1 =0
Substituting the value of C in eq (1) we get
y−2log|y+2|=x+2logx+−2y−x+2=log(y+2)2+logx2y+x−2=log(x2(y+2)2)y−2log⁡|y+2|=x+2log⁡x+−2y−x+2=log(y+2)2+logx2y+x−2=log(x2(y+2)2)
which is the required equation of the curve passing through the point (1, -1)

17. Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Sol: Slope of a tangent is = dydxdydx
Given that ydydx=xydydx=x
Separating the variables the above equation can be written as
y dy=x dxy dy=x dx
Integrating both sides of the equation we get
∫y dy=∫xdxy22=x22+C(1)∫y dy=∫xdx(1)y22=x22+C
 Given that the curve passes throught the point(0, -2)
Substituting x = 0 and y =-2 in eq (1)  we get
(−2)22=022+C42=CC=2(−2)22=022+C42=CC=2
 Substituting the value of C in eq (1) we get
y22=x22+Cy22=x22+2y22=x2+42y2=x2+4y2−x2=4y22=x22+Cy22=x22+2y22=x2+42y2=x2+4y2−x2=4
which is the required equation of the curve passing through  the point (0,-2)

18. At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).

Sol:  The slope of the line passing through the point (x,y) and (-4,-3) is = y+3x+4y+3x+4
We know slope of the tangent =dydxdydx
Given that
dydx=2y+3x+4dyy+4=2(x+3)dxdydx=2y+3x+4dyy+4=2(x+3)dx
Integrating both sides of the equation
∫dyy+3=∫frac2x+4dxlog|y+3|=2log|x+4|+logClog|y+3|=logC(x+4)2y+3=C(x+4)2tag1 ∫dyy+3=∫frac2x+4dxlog⁡|y+3|=2log⁡|x+4|+log⁡Clog⁡|y+3|=log⁡C(x+4)2y+3=C(x+4)2tag1 
The Curve passes through the point (-2,1), Substituting x=-1 and y=1 in eq(1) we get
1+3=C(−2+4)24=C×4C=11+3=C(−2+4)24=C×4C=1
Substituting C=1 in eq (1) we get
y+3=(x+4)2y+3=(x+4)2
Which is the general solution of the given differential equation.

19.  The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Sol: Let r be the radius of the spherical baloon and V be the volume.
Therefore,dVdt=CdVdt=C
dV=CdtdV=Cdt
Integrating both sides we get
∫dV=∫CdtV=Ct+A,where A is a constant4πr33=Ct+A(1)∫dV=∫CdtV=Ct+A,where A is a constant(1)4πr33=Ct+A
 Given that at t =0 , r =3
putting these values in eq (1) we get
4π333=C×0+AA=36π4π333=C×0+AA=36π
equation (1) now becomes
4πr33=Ct+36π(2)(2)4πr33=Ct+36π

When t = 3 sec , r= 6. Putting these values in eq (2) we get
4π633=C×3+36π288π=3C+36πC=84π4π633=C×3+36π288π=3C+36πC=84π
Thus, eq(2) becomes
4πr33=84πt+36π4πr3=3×4π(21t+9)r3=62t+27r=(62t+27)134πr33=84πt+36π4πr3=3×4π(21t+9)r3=62t+27r=(62t+27)13
This is the required  radius of the spherical baloon after t seconds

20.In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years(log 2 = 0.6391)

Sol: Let p be the principal so given that
dpdt=rdpdt=r100pdpp=r100dtdpdt=rdpdt=r100pdpp=r100dt
Integrating both sides of the equation we get
∫dpp=∫r100 dtlog|p|=rt100+logClogp−logC=rt100log|pC|=rt100p=Cert100(1)∫dpp=∫r100 dtlog|p|=rt100+log⁡Clog⁡p−log⁡C=rt100log|pC|=rt100(1)p=Cert100
When t=0, p = 100. Substituting these values in eq (1) we get
100=Ce0C=100100=Ce0C=100
putting this value of C in eq (1) we get
p=100ert100(2)(2)p=100ert100

When t=10, p= 200. Substituting these values in eq (2) we get
200=100er×101002=er10log2=r10r=10log2r=10×0.6931r=6.931200=100er×101002=er10log2=r10r=10log⁡2r=10×0.6931r=6.931
 Thus the value of r is 6.93%

Top 20 Important 100% Common Questions And Answers For H.S Final 2020(Click Here)

Class 12  Ncert  Solution Maths Differential Equations

21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with theis bank, how much will it worth after 10 years(e0.5=1.648)e0.5=1.648).

Sol: Let p be the principal. Given that
dpdt=5100pdpdt=p20dpp=dt20dpdt=5100pdpdt=p20dpp=dt20
Integrating both sides of the equation we get
log|p|=t20+logClog|p|−logC=t20logpC=t20p=Cet20(1)log|p|=t20+log⁡Clog⁡|p|−log⁡C=t20log⁡pC=t20(1)p=Cet20
when t = 0, p= 1000. Substituting these value in the above equation we get
1000=Ce020C=10001000=Ce020C=1000
Putting this value of C in eq(1) we get

p=1000et20(2)(2)p=1000et20
When t =10 eq(2) becomes
p=1000e520p=1000e14p=1000e0.5p=1000×1.648p=1648p=1000e520p=1000e14p=1000e0.5p=1000×1.648p=1648
Thus, after 10 years the deposited amount will become Rs 1648.

22. In a culture, the bacteria count is 1,00,000. The number is increased by 10 % in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Sol: Let the number of bacteria at any time t be n.
Thus given that
dndtα ndndt=kndnn=kdtdndtα ndndt=kndnn=kdt
Integrating the aboce equation we get
log|n|=kt+C(1)(1)log|n|=kt+C
When t=0, n=100000. Substituting these values in eq (1) we get
log100000=k×0+CC=log100000log⁡100000=k×0+CC=log⁡100000
Putting this value of C in eq (1) we get
log|n|=kt+log100000(2)(2)log|n|=kt+log⁡100000
 The number increases by 10 % in 2 hours
So t = 2 and n = 100000 + 10% of 100000
therefore, n = 110000
Substituting these values in eq(2) we get
log110000=2k+log100000log110000−log100000=2klog(110000100000)=2kk=12log(1110)log⁡110000=2k+log⁡100000log⁡110000−log⁡100000=2klog⁡(110000100000)=2kk=12log⁡(1110)
Substituting this value of k in eq (2), we get
log|n|=12log(1110)t+log100000log|n|=12log⁡(1110)t+log⁡100000
Now when n =200000 eq (3) becomes
log200000=12log(1110)t−log100000log200000−log100000=12log(1110)tlog(200000100000)=12log(1110)tlog2=12log(1110)tt=log212log(1110)t=2log2log(1110)log⁡200000=12log⁡(1110)t−log⁡100000log⁡200000−log⁡100000=12log⁡(1110)tlog⁡(200000100000)=12log⁡(1110)tlog⁡2=12log⁡(1110)tt=log212log⁡(1110)t=2log2log⁡(1110)

23.The general solution of the differential equationdydx=ex+ydydx=ex+y is

(A) ex+e−y=Cex+e−y=C                                   (B) ex+ey=Cex+ey=C
(C) e−x+ey=Ce−x+ey=C                               (D) e−x+e−y=Ce−x+e−y=C 

Sol:  The given differential equation can be written as
dyey=ex dxdyey=ex dx
Integrating both sides we get
∫dyey=∫ex dx∫e−y=∫ex dx−e−y+C=exex+e−y=C∫dyey=∫ex dx∫e−y=∫ex dx−e−y+C=exex+e−y=C
So the correct answer is option (A).


(Click Here For Exercise 9.1 & Exercise 9.2 )