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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.2 | Set 2
  • Last Updated : 21 Feb, 2021

Question 11. Assume that a raindrop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the raindrop.

Solution:

Let us considered ‘r’ be the radius of rain drop, volume of the drop be ‘V’ and area of the drop be ‘A’

(dV/dt) proportional to A

(dV/dt) – kA             -(V decreases with increasing in t so negative sing)

Here, k is proportionality constant,



\frac{d(\frac{4π}{3}r^3)}{dt}    = -k(4π r2)

4πr2(dr/dt) = -k(4πr2)

(dr/dt) = -k

Question 12. Find the differential equation of all the parabolas with latus rectum 4a’ and whose axes are parallel to the x-axis.

Solution:

Equation of parabola whose area is parallel to x-axis and vertices at (h, k).

(y – k)2 = 4a(x – h)          -(1)

On differentiating w.r.t x,

2(y – k)(dy/dx) = 4a  



(y – k)(dy/dx) = 2a 

(y-k)=\frac{2a}{\frac{dy}{dx}}            -(2)

Again, differentiating w.r.t x,

d2y/dx2(y – k) + (dy/dx)(dy/dx) = 0   

(\frac{2a}{\frac{dy}{dx}})\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=0

2a(d2y/dx2) + (dy/dx)3 = 0

Question 13. Show that the differential equation of which y=2(x^2-1)+ce^{-x^2} is a solution, is (dy/dx) + 2xy = 4x3

Solution:

y=2(x^2-1)+ce^{-x^2}                -(1)

On differentiating w.r.t x,

\frac{dy}{dx}=4x-2cxe^{-x^2}

On adding 2xy in R.H.S and L.H.S,

\frac{dy}{dx}+2xy=4x-2cxe^{-x^2}+2xy

On putting the value of y in above equation,

\frac{dy}{dx}+2xy=4x-2cxe^{-x^2}+2x(2(x^2-1)+ce^{-x^2})

4x-2cxe^{-x^2}+4x^3-4x+2xce^{-x^2}=4x^3

(dy/dx) + 2xy = 4x3

Question 14. From the differential equation having y = (sin-1x)2 + A cos-1x + B, where A and B are arbitrary constants, as its general solution.

Solution:

y = (sin-1x)2 + A cos-1x + B

On differentiating w.r.t x,

\frac{dy}{dx} = 2sin^{-1}x(\frac{1}{\sqrt{1-x^2}})+A(\frac{-1}{\sqrt{1-x^2}})+0

\sqrt{1-x^2}\frac{dy}{dx}=2sin^{-1}x - A

Again, on differentiating w.r.t x,

\sqrt{1-x^2}\frac{d^2y}{dx^2}+\frac{dy}{dx}\frac{1}{\sqrt{1-x^2}}(-2x)\\ = 2(\frac{1}{\sqrt{1-x^2}})-0\\ =(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}-2=0

Question 15. Form the differential equation of the family of curves represented by the equation (a being the parameter)

(i) (2x + a)2 + y2 = a2  

Solution:

(2x + a)2 + y2 = a2          -(1)

On differentiating w.r.t x,

2(2x + a) + 2y(dy/dx) = 0 

(2x + a) + y(dy/dx) = 0 

a = -2x – y(dy/dx)          -(2)

On putting the value of ‘a’ in eq(1), we have

(2x-2x-y\frac{dy}{dx})^2+y^2=(-2x-y\frac{dy}{dx})^2

(y\frac{dy}{dx})^2+y^2=(4x^2+4xy\frac{dy}{dx}+y^2(\frac{dy}{dx})^2)

y2 = 4x2 + 4xy(dy/dx)

y2 – 4x2 – 4xy(dy/dx) = 0

(ii) (2x – a)2 – y2 = a2      

Solution:

(2x – a)2 – y2 = a2          

4x2 – 4ax + a2 – y2 = a

4ax = 4x2 – y2

a = (4x2 – y2)/4x        

On differentiating w.r.t x,

[\frac{4x(8x-2y\frac{dy}{dx}-4(4x^2-y^2}{(4x)^2}]=0

32x^2-8xy\frac{dy}{dx}-16x^2+4y^2=0

16x^2+4y^2-8xy\frac{dy}{dx}=0

4x2 + y2 = 2xy(dy/dx)

(iii) (x – a)2 + 2y2 = a        

Solution:

(x – a)2 + 2y2 = a2          -(1)

On differentiating w.r.t x,

2(x – a) + 4y(dy/dx) = 0

(x – a) + 2y(dy/dx) = 0

a = x + 2y(dy/dx)          -(2)

On putting the value of a in eq(1)

(x-x-2y\frac{dy}{dx})^2+2y^2=(x+2y\frac{dy}{dx})^2

4y^2(\frac{dy}{dx})^2+2y^2=x^2+4xy(\frac{dy}{dx})+4y^2(\frac{dy}{dx})^2

2y2 – 4xy(dy/dx) – x2 = 0

Question 16. Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

(i) x2 + y2 = a2  

Solution:

x2 + y2 = a2

On differentiating w.r.t x,

2x + 2y(dy/dx) = 0

x + y(dy/dx) = 0

(ii) x2 – y2 = a2  

Solution:

x2 – y2 = a2 

On differentiating w.r.t x,

2x – 2y(dy/dx) = 0

x – y(dy/dx) = 0

(iii) y2 = 4ax 

Solution:

y2 = 4ax 

(y2/x) = 4a

On differentiating w.r.t x,

[\frac{2xy\frac{dy}{dx}-y^2}{x^2}]=0

2xy(dy/dx) – y2 = 0

2x(dy/dx) – y = 0

(iv) x2 + (y – b)2 = 1   

Solution:

x2 + (y – b)2 = 1          -(1)

On differentiating w.r.t x,

2x + 2(y – b)(dy/dx) = 0

(y-b)=-\frac{x}{\frac{dy}{dx}}

On putting the value of (y – b) in eq(1)

x^2+[-\frac{x}{\frac{dy}{dx}}]^2=1

x2(dy/dx)2 + x2 = (dy/dx)2

x2[(dy/dx)2 + 1] = (dy/dx)2

(v) (x – a)2 – y2 = 1    

Solution:

(x – a)2 – y2 = 1           -(1)

On differentiating w.r.t x,

2(x – a) – 2y(dy/dx) = 0

(x – a) – y(dy/dx) = 0

(x – a) = y(dy/dx)

On putting the value of (y – b) in eq(i), we get

y2(dy/dx)2 – y2 = 1

y2[(dy/dx)2 – 1] = 1

(vi)\frac{x^2}{a^2}-\frac{y^2}{b^2}=1  

Solution:

We have,

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1            -(1)

\frac{b^2x^2-a^2y^2}{a^2b^2}=1

{(bx)2 – (ay)2} = (ab)2           -(2)

On differentiating w.r.t x,

2xb2 – 2a2y(dy/dx) = 0

xb2 – a2y(dy/dx) = 0           -(3)

Again, differentiating w.r.t x,

b^2-a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]

b^2=a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]

On putting the value of b2 in equation(3), we get

xb2 – a2y(dy/dx) = 0  

xa^2[y(\frac{dy}{dx}^2)+y\frac{d^2y}{dx^2}]-a^2y\frac{dy}{dx}=0

xy\frac{d^2y}{dx^2}+x(\frac{dy}{dx})^2-y(\frac{dy}{dx})=0

x[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2]=y(\frac{dy}{dx})

(vii) y2 = 4a(x – b) 

Solution:

We have,

y2 = 4a(x – b)          

On differentiating w.r.t x,

2y(dy/dx) = 4a

Again differentiating w.r.t x,

2[(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]=0

[(dy/dx)2 + y(d2y/dx2)] = 0

(viii) y = ax3

Solution:

We have,

y = ax3         -(1)

On differentiating w.r.t x,

(dy/dx) = 3ax2

From eq(1),

a=(y/x3         -(1)

On putting the value of a in eq(1)

dy/dx = 3(y/x3) × x2

x(dy/dx) = 3y

(ix) x2 + y2 = ax3 

Solution:

We have,

x2 + y2 = ax3       

a = (x2 + y2)/(x3

On differentiating w.r.t x,

\frac{(2x+2y\frac{dy}{dx})(x^3)-(3x^2)(x^2+y^2)}{(x^3)^2}=0

2x^3y\frac{dy}{dx}+2x^4-3x^4-3x^2y^2=0

2x3y(dy/dx) = x4 + 3x2y2

2x3y(dy/dx) = x2(x2 + 3y2)

2xy(dy/dx) = (x2 + 3y2)

(x) y = eax

Solution:

We have,

y = eax         -(1)

On differentiating w.r.t x,

dy/dx = aeax

dy/dx = ay         -(2)

y = eax 

On taking log both side, we get

logy = ax

a = (logy/x)

Now, put the value of ‘a’ in eq(2)

(dy/dx) = logy/x) × y

x(dy/dx) = ylogy

Question 17. Form the differential equation representing the family of ellipses having foci on the x-axis and the centre at the origin.

Solution:

We have,

Equation of ellipse having foci on the x-axis,

 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1            -(where a > b)

\frac{b^2x^2+a^2y^2}{a^2b^2}=1

(bx)2 + (ay)2 = (ab)2         -(1)

On differentiating above equation w.r.t x,

2b2x + 2a2y(dy/dx) = 0

b2x + a2y(dy/dx) = 0         -(2)

Again, differentiating w.r.t x,

b^2+a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]=0

b^2=-a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]

On putting the value of b2 in eq(2),

xb2 + a2y(dy/dx) = 0

-xa^2[y(\frac{dy}{dx}^2)+y\frac{d^2y}{dx^2}]+a^2y\frac{dy}{dx}=0

xy\frac{d^2y}{dx^2}+x(\frac{dy}{dx})^2-y(\frac{dy}{dx})=0

x[y(d2y/dx2) + (dy/dx)2] = y(dy/dx)

Question 18. Form the differential equation of the family of hyperbolas having foci on the X-axis and center at the origin

Solution:

We have,

Equation of a hyperbola having a Centre at the origin and foci along x-axis

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1           -(1)

\frac{b^2x^2-a^2y^2}{a^2b^2}=1

(bx)2-(ay)2=(ab)2         -(2)

On differentiating above equation w.r.t x,

2xb2-2a2y(dy/dx)=0

xb2-a2y(dy/dx)=0          -(3)

Again, differentiating above equation w.r.t x,

b^2-a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]=0

b^2=a^2[y(\frac{dy}{dx})(\frac{dy}{dx})+y\frac{d^2y}{dx^2}]

Putting the value of b2 in equation(3),

xb^2-a^2y\frac{dy}{dx}=0

xa^2[y(\frac{dy}{dx}^2)+y\frac{d^2y}{dx^2}]-a^2y\frac{dy}{dx}=0

xy(d2y/dx2) + x(dy/dx)2 – y(dy/dx) = 0

x[y(d2y/dx2) +(dy/dx)2] = y(dy/dx)

This is required differential equation.

Question 19. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axis. 

Solution:

We have,

Let (-a, a) be the coordinates of the center of circle 

So, the equation of circle is given by,

(x + a)2 + (y – b)2 = a2         -(1)

x2 + 2ax + a2 + y2 – 2ay + a2 = 0         -(2)

On differentiating above equation w.r.t x,

2x + 2a + 2y(dy/dx) – 2a(dy/dx) = 0

x + a + y(dy/dx) – a(dy/dx) = 0

x+y\frac{dy}{dx}=a(\frac{dy}{dx}-1)

a=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}

On substituting the value of ‘a’ in eq(2)

[x+\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}]^2+[y-\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}]^2=[\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}-1}]^2

Let, (dy/dx) = p 

[xp – x + x + yp]2 + [yp – y – x – yp]2 = [x + yp]2

(x + y)2p2 + (x + y)2 = (x + yp)2

(x + y)2[p2 + 1] = (x + yp)2          -(where (dy/dx) = p) 

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