12th NCERT Differential Equation Exercise 9.2 Number of questions 12

Hi

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation

Question (1)

y = e^x + 1 : y'' - y' = 0

Solution

y = e^x + 1
differentiate w.r.t to x
y' = e^x
differentiate w.r.t to x
y'' = e^x
y'' = y'
∴ y'' - y' = 0

Question (2)

y=x² + 2x + C : y' -2x - 2 = 0

Solution

y = x² + 2x + c
differentiate w.r.t to x
\[\frac{{dy}}{{dx}} = 2x + 2\] ∴ y' - 2x -2 = 0

Question (3)

y=cosx + C : y' + sinx = 0

Solution

y = cosx + C
Differentiate w.r.t x
\[\frac{{dy}}{{dx}} = - \sin x\] y' + sinx =0

Question (4)

\[y = \sqrt {1 + {x^2}} \;:\;y' = \frac{{xy}}{{1 + {x^2}}}\]

Solution

\[y = \sqrt {1 + {x^2}} \] Differentiate w.r.t x
\[\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {1 + {x^2}} }} \cdot \frac{d}{{dx}}\left( {1 + {x^2}} \right)\] \[y' = \frac{1}{{\require{cancel}\cancel{2}\sqrt {1 + {x^2}} }} \cdot \cancel{2}x\] \[y' = \frac{x}{{\sqrt {1 + {x^2}} }}\] \[y' = \frac{{x\sqrt {1 + {x^2}} }}{{1 + {x^2}}}\] \[y' = \frac{{xy}}{{1 + {x^2}}}\]

Question (5)

y = Ax : xy' = y (x ≠ 0)

Solution

y = Ax
Differentiate w.r.t x
\[\frac{{dy}}{{dx}} = A\] y'=A
multiply by x
xy' = Ax
xy' = y

Question (6)

\[\begin{array}{l}y = x\sin x\;:\;xy' = y + x\sqrt {{x^2} - {y^2}} \\\quad \quad \quad \quad \;(x \ne 0{\kern 1pt} \,and\,x > y{\kern 1pt} \,or\,x < - y)\end{array}\]

Solution

y= xsinx
Differentiate w.r.t x
\[y' = x\frac{d}{{dx}}\sin x + \sin x\frac{d}{{dx}}x\] y'= xcosx + sinx
multiply by x
∴ xy' = x²cosx + xsinx
xy' = x²cosx + y
xy'=x(xcosx) + y
\[xy' = x\sqrt {{x^2}{{\cos }^2}x} + y\] \[xy' = x\sqrt {{x^2}\left( {1 - {{\sin }^2}x} \right)} + y\] \[xy' = x\sqrt {{x^2} - {x^2}{{\sin }^2}x} + y\] \[xy' = x\sqrt {{x^2} - {y^2}} + y\]

Question (7)

\[xy = \log y + C\;\quad :\quad y' = \frac{{{y^2}}}{{1 - xy}}\left( {xy \ne 1} \right)\]

Solution

xy = logy + C
Differentiate w.r.t x
\[x\frac{{dy}}{{dx}} + y\left( 1 \right) = \frac{1}{y}\frac{{dy}}{{dx}}\] \[x\frac{{dy}}{{dx}} - \frac{1}{y}\frac{{dy}}{{dx}} = - y\] \[\left( {\frac{{xy - 1}}{y}} \right)\frac{{dy}}{{dx}} = - y\] \[y' = \frac{{ - {y^2}}}{{xy - 1}}\] \[y' = \frac{{{y^2}}}{{1 - xy}}\]

Question (8)

y-cosy = x : (ysiny + cosy +x)y' = y

Solution

y-cosy = x
Differentiate w.r.t x
\[y' - \sin y \cdot \frac{{dy}}{{dx}} = 1\] \[y' - \sin y \cdot y' = 1\] multiply by y
yy' + ysiny. y' = y
y'(y + ysiny) = y
y'(x+cosy + ysiny) = y
∴ (ysiny + cosy + x) y' = y

Question (9)

x + y = tan^-1y : y²y' + y² + 1 = 0

Solution

x + y = tan^-1y
Differentiate w.r.t x
\[1 + \frac{{dy}}{{dx}} = \frac{1}{{1 + {y^2}}} \cdot \frac{{dy}}{{dx}}\] \[1 + y' = \frac{{y'}}{{1 + {y^2}}}\] (1+y²) (1+y') = y'
\[1 + \require{cancel}\cancel{y'} + {y^2} + {y^2}y' = \cancel{y'}\] y²y'+ y² + 1 = 0

Question (10)

\[y = \sqrt {{a^2} - {x^2}} \;x \in \left( { - a,a} \right):x + y\frac{{dy}}{{dx}} = 0(y \ne 0)\]

Solution

\[y = \sqrt {{a^2} - {x^2}} \;x \in \left( { - a,a} \right)\] Differentiate w.r.t x
\[\frac{{dy}}{{dx}} = \frac{1}{{\cancel{2}\sqrt {{a^2} - {x^2}} }} \times \left( { - \cancel{2}x} \right)\] \[y' = \frac{{ - x}}{y}\] yy' = -x
x + yy' = 0

Question (11)

The number of arbitrary constants in the general solution of a differential equation of fourth order are
(A) 0   (B) 2   (C) 3   (D) 4

Solution

Number of arbitrary constant of 4th order are 4
as number of arbitrary constant = degree

Question (12)

The number of arbitrary constants in the particular solution of a differential equation of third order are
(A) 3   (B) 2   (C) 1   (D) 0

Solution

Number of arbitrary constant in perticular solution = 0