12th NCERT Differential Equation Exercise 9.3 Number of questions 12
Hi
In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constant a and b

Question (1)

$\frac{x}{a} + \frac{y}{b} = 1$

Solution

$\frac{x}{a} + \frac{y}{b} = 1$ Here as we have two arbitrary constant we can take up to 2nd order
differentiating w.r.t x $\frac{1}{a} + \frac{1}{b}y' = 0$ $\frac{1}{b}y' = \frac{{ - 1}}{a}$ $y' = \frac{{ - b}}{a}$ differentiating w.r.t x again y2 = 0

Question (2)

y2 = a(b2 -x2)

Solution

As two constant, find upto 2nd order to get differential equation where and b are not present
differentiating w.r.t x
$2y\frac{{dy}}{{dx}} = a\left( {0 - 2x} \right)$
∴ yy1 = -ax
differentiating w.r.t x again
yy2 + y1 y1 = -a
Multiply by x on both side
xyy2 + x y12 = -ax
xyy2 + x y12 = yy1
xyy2 + x y12 - yy1 = 0

Question (3)

y = ae3x + b e-2x

Solution

differentiating w.r.t x
y' =a e3x(3) + be-2x (-2)
y' = 3ae3x - 2be-2x
y' = 3[ ae3x + be-2x] -5be-2x
y' = 3y -5be-2x
y' - 3y = -5be-2x _ _ _ (1)
differentiating w.r.t x
y2 - 3y1 = -5be-2x(-2)
y2 - 3y1 = -2[y'-3y])
y2 - 3y1 +2y1-6y=0
y2 -y1 - 6y = 0

Question (4)

y = e2x (a + bx)

Solution

differentiating w.r.t x
${y_1} = {e^{2x}}\frac{d}{{dx}}\left( {a + bx} \right) + \left( {a + bx} \right)\frac{d}{{dx}}{e^{2x}}$ ${y_1} = {e^{2x}}b + \left( {a + bx} \right){e^{2x}} \cdot \left( 2 \right)$ y1 = be2x + 2y
y1 - 2y = be2x _ _ _ (1) differentiating w.r.t x
y2 - 2y1 = be2x.(2)
Replacing value of be2x from (1)
y2 - 2y1 = 2 (y1 -2y)
y2 - 2y1 = 2y1 - 4y
y2 - 4y1 + 4y = 0

Question (5)

y = ex (a cos x + b sin x)

Solution

differentiating w.r.t x
y1 = ex [ a (-sinx) + bcos x] + (acosx + bsinx) ex
y1 = ex ( -asin x + bcs x ) + y
y1 - y = ex ( -asin x + bcos x )
differentiating w.r.t x
y2 - y1 = ex [ -acosx + b(-sinx)] + (-asinx + bcosx) ex
y2 - y1 = -ex ( acos x + bsiinx) + (-asin x + bcosx) ex
y2 - y1 = -y + (-asinx + bcosx) ex
y2 - y1 = -y + y1 - y
y2- y1 +y - y1 + y = 0
y2 -2y1 + 2y = 0

Question (6)

Form the differential equation of the family of circles touching the y-axis at origin

Solution

Form differential equation of family of circle touching y-axis at origin
As O touches y-axis at (0, 0) it centre lines on x-axis.
Let radius be "r" ten centre (r, 0)
[ Equation 9.3 ] Equations of circle is
(x-r)2 + y 2 = r2
x2 - 2xr + r2 + y2 = r2
x2 + y2 - 2xr = 0 _ _ _ (1)
differentiating w.r.t x
$2x + 2y\frac{{dy}}{{dx}} - 2r = 0$ x + yy1 = r
Replacing value of 'r' in equation (1) we get
x2 + y2 - 2x(x + yy1) = 0
x2 + y2 -2x2 - 2xyy1 = 0
y2 - x2 - 2xyy1 = 0
2xyy1 + x2 = y2

Question (7)

Form the differential equation of the family of parabolas havig vertex at origin and axis along positive y-axis

Solution

Form the differential equation of the family of parabolas havig vertex at (0, 0) along positive y-axis.
[Diagram d7] The standard equation of parabola is
x2 = 4ay _ _ _ (1) differentiating w.r.t x
2x = 4ay1
x = 2ay1
x/2y1 = a
Replacing value of a in equation (1)
x2 = 4ay
${x^2} =\require{cancel} \cancel{4}^2\left( {\frac{x}{{\cancel{2}{y_1}}}} \right)y$ x2 y1 = 2xy
xy1 = 2y
xy1 - 2y = 0

Question (8)

Form the differential equation of the family of ellipses having foci on y-axis and centre at origin

Solution

Family of ellipse focus on y-axis
[Diagram 8 ] $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ ∴ b2x2 + a2y2 = a2 b2
differentiating w.r.t x
$2x{b^2} + 2y{a^2}\frac{{dy}}{{dx}} = 0$ b2x + a2 y y1 = 0 $y{y_1} = \frac{{ - {b^2}x}}{{{a^2}}}$ differentiating w.r.t x again
$y{y_2} + {y_1} \cdot \left( {{y_1}} \right) = \frac{{ - {b^2}}}{{{a^2}}}$ Multiply by x on both side
$xy{y_2} + xy_1^2 = \frac{{ - {b^2}x}}{{{a^2}}}$ xyy2 + x y12 = yy1 xyy2 + x y12 - yy1 = 0

Question (9)

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin

Solution

[Image D-9] $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ differentiating w.r.t x
$\frac{{2x}}{{{a^2}}} - \frac{{2y}}{{{b^2}}}{y_1} = 0$ $\frac{{2x}}{{{a^2}}} = \frac{{2y}}{{{b^2}}}{y_1}$ $\frac{{y{y_1}}}{x} = \frac{{{b^2}}}{{{a^2}}}$ differentiating w.r.t x
$\frac{{x\left[ {y{y_2} + {y_1}\left( {{y_1}} \right)} \right] - y{y_1}}}{{{x^2}}} = 0$ ∴ xyy2 + x y12 - yy1 = 0

Question (10)

Form the differential equation of the family oof circles having centre on y-axis and radius 3 units.

Solution

Let centre C (0, a)
(x-0)2 + (y-a)2 = 9 ..... (1)
x2 + y2 - 2axy + a2 - 9 = 0
differentiating w.r.t x
2x+2yy1 - 2ay1 = 0
x + yy1 = ay1
$a = \frac{{x + y{y_1}}}{{{y_1}}}$ ${x^2} + {y^2} - 2y\left( {\frac{{x + y{y_1}}}{{{y_1}}}} \right) + {\left( {\frac{{x + y{y_1}}}{{{y_1}}}} \right)^2} - 9 = 0$ ${x^2} + {y^2} - 2y\left( {\frac{{x + y{y_1}}}{{{y_1}}}} \right) + \frac{{{x^2} + 2xy{y_1} + {y^2}y_1^2}}{{y_1^2}} - 9 = 0$ x2y12+y2y12-2yy1(x+yy1) +x2 +2xyy1+y2y12 -9y12 = 0
(x2 - a)y12+y2y12 -2(x+yy1)yy1 +(x+yy1)2 = 0
(x2-9)y12+(x+{yy1}-{yy1})2
(x2-9)y12 + x2 =0

Question (11)

Which of the following differential equations has y = c1ex + c2e-x $\left( A \right)\frac{{{d^2}y}}{{d{x^2}}} + y = 0$ $\left( B \right)\frac{{{d^2}y}}{{d{x^2}}} - y = 0$ $\left( C \right)\frac{{{d^2}y}}{{d{x^2}}} + 1 = 0$ $\left( D \right)\frac{{{d^2}y}}{{d{x^2}}} - 1 = 0$

Solution

y = C1 ex + C2e-x
differentiating w.r.t x
y1 = C1ex + C2 e-x(-1)
y1 = C1ex - C2 e-x
y1 = (C1ex + C2 e-x) - 2C2e-x
y1 - y = - 2C2e-x
differentiating w.r.t x
y2 - y1 = - 2C2e-x(-1)
y2 - y1 = -(y1 - y)
y2 - {y1} = - {y1} + y
y2 = y
y2 - y = 0
Option (B) correct

Question (12)

Which of the following differential equations has y=x as one of its particular solution? $\left( A \right)\frac{{{d^2}y}}{{d{x^2}}} - {x^2}\frac{{dy}}{{dx}} + xy = x$ $\left( B \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + xy = x$ $\left( C \right)\frac{{{d^2}y}}{{d{x^2}}} - {x^2}\frac{{dy}}{{dx}} + xy = 0$ $\left( D \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + xy = 0$

Solution

y = x
differentiating w.r.t x
$\frac{{dy}}{{dx}} = 1$ differentiating w.r.t x
$\frac{{{d^2}y}}{{d{x^2}}} = 0$ Substituting value
Option (C) is correct