12th NCERT Differential Equation Exercise 9.3 Number of questions 12

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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

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 = ae^3x + b e-2x

Solution

differentiating w.r.t x
y' =a e^3x(3) + be^-2x (-2)
y' = 3ae^3x - 2be^-2x
y' = 3[ ae^3x + 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 = e^2x (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 = be^2x + 2y
y₁ - 2y = be^2x _ _ _ (1) differentiating w.r.t x
y₂ - 2y₁ = be^2x.(2)
Replacing value of be^2x from (1)
y₂ - 2y₁ = 2 (y₁ -2y)
y₂ - 2y₁ = 2y₁ - 4y
y₂ - 4y₁ + 4y = 0

Question (5)

y = e^x (a cos x + b sin x)

Solution

differentiating w.r.t x
y₁ = e^x [ a (-sinx) + bcos x] + (acosx + bsinx) e^x
y₁ = e^x ( -asin x + bcs x ) + y
y₁ - y = e^x ( -asin x + bcos x )
differentiating w.r.t x
y₂ - y₁ = e^x [ -acosx + b(-sinx)] + (-asinx + bcosx) e^x
y₂ - y₁ = -e^x ( acos x + bsiinx) + (-asin x + bcosx) e^x
y₂ - y₁ = -y + (-asinx + bcosx) e^x
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 = c₁e^x + 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 e^x + C₂e^-x
differentiating w.r.t x
y₁ = C₁e^x + C₂ e^-x(-1)
y₁ = C₁e^x - C₂ e^-x
y₁ = (C₁e^x + 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