Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.2 | Set 2

Last Updated : 10 May, 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= a²
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= a²

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 = e^ax

Solution:

We have,
y = e^ax -(1)
On differentiating w.r.t x,
dy/dx = ae^ax
dy/dx = ay -(2)
y = e^ax
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 centre 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 centre 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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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.2 | Set 2

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.

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

Question 13. Show that the differential equation of which is a solution, is (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.

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

(ii) (2x – a)2 – y2 = a2

(iii) (x – a)2 + 2y2 = a2

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

(i) x2 + y2 = a2

(ii) x2 – y2 = a2

(iii) y2 = 4ax

(iv) x2 + (y – b)2 = 1

(v) (x – a)2 – y2 = 1

(vi)

(vii) y2 = 4a(x – b)

(viii) y = ax3

(ix) x2 + y2 = ax3

(x) y = eax

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

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

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

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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

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.

(i) (2x + a)²+ y²= a²

(ii) (2x – a)²– y²= a²

(iii) (x – a)²+ 2y²= a²

(i) x²+ y²= a²

(ii) x²– y²= a²

(iii) y²= 4ax

(iv) x²+ (y – b)²= 1

(v) (x – a)²– y²= 1

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

(vii) y²= 4a(x – b)

(viii) y = ax³

(ix) x²+ y²= ax³

(x) y = e^ax