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 = 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 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)
Summary
Exercise 22.2 | Set 2 focuses on solving first-order differential equations, particularly those that can be classified as homogeneous equations. Key concepts covered include:
- Homogeneous differential equations
- Method of substitution (y = vx)
- Separation of variables
- Particular solutions with initial conditions
- Orthogonal trajectories