12th NCERT Differential Equation Exercise 9.6 Number of questions 19
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For each of the differential equations given in Exercises 1 to 12, find the general solution

Question (1)

$\frac{{dy}}{{dx}} + 2y = \sin x$

Solution

Compare given equation with standard equation $\frac{{dy}}{{dx}} + py = q\left( x \right)$ We get p =2 $M.F. = {e^{\int {pdx} }}$ $M.F. = {e^{\int {2dx} }}$ $M.F. = {e^{2x}}$ $\therefore {e^{2x}}\frac{{dy}}{{dx}} + 2y{e^{2x}} = \sin x \cdot {e^{2x}}$ $\therefore \frac{d}{{dx}}\left( {y{e^{2x}}} \right) = \sin x \cdot {e^{2x}}$ $\int {\frac{d}{{dx}}} \left( {y{e^{2x}}} \right)dx = \int {\sin x} \cdot {e^{2x}}dx$ $y{e^{2x}} = \int {\sin x} \cdot {e^{2x}}dx$ $y{e^{2x}} = I$ $\text {where} \qquad I = \int {\sin x} \cdot {e^{2x}}dx$
Here u = sinx and v=e2x
Using integration by parts
$I = \sin x\int {{e^{2x}}dx} - \int {\left[ {\frac{d}{{dx}}\sin x\int {{e^{2x}}dx} } \right]dx}$ $I = \sin x\frac{{{e^{2x}}}}{2} - \int {\cos x} \frac{{{e^{2x}}}}{2}dx$ $I = \sin x\frac{{{e^{2x}}}}{2} - \frac{1}{2}\int {\cos x} \cdot {e^{2x}}dx$ Using integration by parts $I = \sin x\frac{{{e^{2x}}}}{2} - \frac{1}{2}\left[ {\cos x\int {{e^{2x}}dx} - \int {\left( {\frac{d}{{dx}}\cos x\int {{e^{2x}}dx} } \right)dx} } \right]$ $I = \sin x\frac{{{e^{2x}}}}{2} - \frac{1}{2}\left[ {\cos x\frac{{{e^{2x}}}}{2} - \int { - \sin xdx\frac{{{e^{2x}}}}{2}} } \right]dx$ $I = \sin x\frac{{{e^{2x}}}}{2} - \cos x\frac{{{e^{2x}}}}{4} - \frac{1}{4}\left[ {\int {\sin x \cdot {e^{2x}}} } \right]dx$ $I = \frac{{{e^{2x}}}}{4}\left[ {2\sin x - \cos x} \right] - \frac{1}{4}I$ $\frac{5}{4}I = \frac{{{e^{2x}}}}{4}\left[ {2\sin x - \cos x} \right] + c$ $I = \frac{{{e^{2x}}}}{5}\left[ {2\sin x - \cos x} \right] + c$ ye2x = I
$y{e^{2x}} = \frac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right) + c$ $y = \frac{1}{5}\left( {2\sin x - \cos x} \right) + \frac{c}{{{e^{2x}}}}$ $y = \frac{{2\sin x - \cos x}}{5} + c{e^{ - 2x}}$

Question (2)

$\frac{{dy}}{{dx}} + 3y = {e^{ - 2x}}$

Solution

p =3 $I.F. = {e^{\int {pdx} }}$ $I.F. = {e^{\int {3dx} }}$ $I.F. = {e^{3x}}$ Multiply e3x to equation ${e^{3x}}\frac{{dy}}{{dx}} + 3y{e^{3x}} = {e^{ - 2x}} \cdot {e^{3x}}$ $\therefore \frac{{dy}}{{dx}}\left( {y{e^{3x}}} \right) = {e^x}$ $\int {\frac{{dy}}{{dx}}\left( {y{e^{3x}}} \right)} dx = \int {{e^x}dx}$ ye3x = ex + c $y = \frac{{{e^x}}}{{{e^{3x}}}} + \frac{c}{{{e^{3x}}}}$ $y = {e^{ - 2x}} + c{e^{ - 3x}}$

Question (3)

$\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}$

Solution

p=1/x $I.F. = {e^{\int {pdx} }}$ $I.F. = {e^{\int {\frac{1}{x}dx} }}$ $I.F. = {e^{\log x}}$ $I.F. = {x^{\log e}}$ I.F. = x $x\frac{{dy}}{{dx}} + y = {x^3}$ $\frac{d}{{dx}}\left( {yx} \right) = {x^3}$ $\int {\frac{d}{{dx}}\left( {yx} \right)} dx = \int {{x^3}dx}$ $yx = \frac{{{x^4}}}{4} + c$ $y = \frac{{{x^3}}}{4} + \frac{c}{x}$

Question (4)

$\frac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x\quad \left( {0 \le x \le \frac{\pi }{2}} \right)$

Solution

p=secx $I.F. = {e^{\int {pdx} }}$ $I.F. = {e^{\int {\sec xdx} }}$ $I.F. = {e^{\log \left( {\sec x + \tan x} \right)}}$ I.F. = secx + tanx
Multiplying by integral factor
$\left( {\sec x + \tan x} \right)\frac{{dy}}{{dx}} + \sec x\left( {\sec x + \tan x} \right)y = \tan x\left( {\sec x + \tan x} \right)$ $\frac{d}{{dx}}\left( {\sec x + \tan x} \right)y = \sec x\tan x + {\tan ^2}x$ $\int {\frac{d}{{dx}}\left( {\sec x + \tan x} \right)ydx} = \int {\left( {\sec x\tan x + {{\tan }^2}x} \right)dx}$ $\left( {\sec x + \tan x} \right)y = \int {\sec x\tan xdx + \int {{{\tan }^2}xdx} }$ $\left( {\sec x + \tan x} \right)y = \sec x + \int {\left( {{{\sec }^2}x - 1} \right)dx}$ $\left( {\sec x + \tan x} \right)y = \sec x + \tan x - x + c$ $y = 1 - \frac{x}{{\left( {\sec x + \tan x} \right)}} + \frac{c}{{\left( {\sec x + \tan x} \right)}}$

Question (5)

${\cos ^2}x\frac{{dy}}{{dx}} + y = \tan x\quad \left( {0 \le x \le \frac{\pi }{2}} \right)$

Solution

$\frac{{dy}}{{dx}} + \frac{y}{{{{\cos }^2}x}} = \frac{{\tan x}}{{{{\cos }^2}x}}$ comparing to standard equation p=sec2x $IF = {e^{\int {pdx} }}$ $IF = {e^{\int {{{\sec }^2}xdx} }}$ $IF = {e^{\tan x}}$ Multiply by etanx ${e^{\tan x}}\frac{{dy}}{{dx}} + y{e^{\tan x}}{\sec ^2}x = {e^{\tan x}}\tan x{\sec ^2}x$ $\frac{d}{{dx}}\left( {y{e^{\tan x}}} \right) = {e^{\tan x}}\tan x{\sec ^2}x$ $\int {\frac{d}{{dx}}\left( {y{e^{\tan x}}} \right)dx} = \int {{e^{\tan x}}\tan x{{\sec }^2}xdx}$ $y{e^{\tan x}} = \int {{e^{\tan x}}\tan x{{\sec }^2}xdx}$ Let tanx = t
sec2xdx = dt $y{e^{\tan x}} = \int {{e^t}{\mathop{\rm t}\nolimits} dt}$ $y{e^{\tan x}} = \int {{\mathop{\rm t}\nolimits} {e^t}dt}$ $y{e^{\tan x}} = t\int {{e^t}dt} - \int {\left( {\frac{d}{{dt}}t\int {{e^t}} dt} \right)dt}$ $y{e^{\tan x}} = t{e^t} - \int {1{e^t}dt}$ $y{e^{\tan x}} = t{e^t} - {e^t} + c$ $y{e^{\tan x}} = \tan x{e^{\tan x}} - {e^{\tan x}} + c$ $y = \tan x - 1 + c{e^{ - \tan x}}$

Question (6)

$x\frac{{dy}}{{dx}} + 2y = {x^2}\log x$

Solution

$\frac{{dy}}{{dx}} + \frac{{2y}}{x} = x\log x - - - (1)$ p = 2/x $IF = {e^{\int {pdx} }}$ $IF = {e^{\int {\frac{2}{x}dx} }}$ $IF = {e^{2\log x}}$ $IF = {x^{2\log e}}$ IF=x2
Multiply by x2 to equation (1)
${x^2}\frac{{dy}}{{dx}} + 2xy = {x^3}\log x$ $\frac{d}{{dx}}\left( {{x^2}y} \right) = {x^3}\log x$ $\int {\frac{d}{{dx}}\left( {{x^2}y} \right)dx} = \int {{x^3}\log xdx}$ logx = u and x3 =v $\int {\frac{d}{{dx}}\left( {{x^2}y} \right)dx} = \int {\log x \cdot {x^3}dx}$ ${x^2}y = \log x\int {{x^3}dx - \int {\left( {\frac{d}{{dx}}\log x\int {{x^3}dx} } \right)dx} }$ ${x^2}y = \log x \cdot \frac{{{x^4}}}{4} - \int {\frac{1}{x} \cdot \frac{{{x^4}}}{4}dx}$ ${x^2}y = \log x \cdot \frac{{{x^4}}}{4} - \frac{1}{4} \cdot \frac{{{x^4}}}{4} + c$ $y = \log x \cdot \frac{{{x^2}}}{4} - \frac{{{x^2}}}{{16}} + c{x^{ - 2}}$

Question (7)

$x\log x\frac{{dy}}{{dx}} + y = \frac{2}{x}\log x$

Solution

$\frac{{dy}}{{dx}} + \frac{y}{{x\log x}} = \frac{2}{{{x^2}}} - - - (1)$ $p = \frac{1}{{x\log x}}$ $I.F. = {e^{pdx}}$ $I.F. = {e^{\int {\frac{1}{{x\log x}}} dx}}$ log x = t
(1/x)dx = dt
$I.F. = {e^{\int {\frac{{dt}}{t}} }}$ $I.F. = {e^{\log t}}$ I.F. = t = logx
Multiply by logx equation (1)
$\log x\frac{{dy}}{{dx}} + \frac{y}{x} = \frac{{2\log x}}{{{x^2}}}$ $\therefore \frac{d}{{dx}}\left( {\log x \cdot y} \right) = \frac{{2\log x}}{{{x^2}}}$ $\int {\frac{d}{{dx}}\left( {y\log x} \right)dx} = \int {\frac{{2\log x}}{{{x^2}}}dx}$ $y\log x = 2\left[ {\int {\log x \cdot {x^{ - 2}}dx} } \right]$ $y\log x = 2\left[ {\log x\int {{x^{ - 2}}dx - \int {\left( {\frac{d}{{dx}}\left( {\log x} \right)\int {{x^{ - 2}}dx} } \right)dx} } } \right]$ $y\log x = 2\left[ {\log x\frac{{{x^{ - 1}}}}{{ - 1}} - \int {\frac{1}{x} \times \frac{{{x^{ - 1}}}}{{ - 1}}dx} } \right]$ $y\log x = \frac{{ - 2\log x}}{x} + 2\int {{x^{ - 2}}dx}$ $y\log x = \frac{{ - 2\log x}}{x} + \frac{{2{x^{ - 1}}}}{{ - 1}} + c$ $y\log x = \frac{{ - 2\log x}}{x} - \frac{2}{x} + c$ $y = \frac{{ - 2}}{x} - \frac{2}{{x\log x}} + \frac{c}{{\log x}}$

Question (8)

(1+x2)dy+2xy dx = cot x dx ( x ≠ 0)

Solution

(1+x2)dy= (cotx-2xy)dx $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = \cot x - 2xy$ $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = \cot x$ $\frac{{dy}}{{dx}} + \frac{{2xy}}{{1 + {x^2}}} = \frac{{\cot x}}{{1 + {x^2}}} - - - (1)$ comparing we get $p = \frac{{2x}}{{1 + {x^2}}}$ $IF = {e^{\int {pdx} }}$ $IF = {e^{\int {\frac{{2x}}{{1 + {x^2}}}dx} }}$ $IF = {e^{\log \left( {1 + {x^2}} \right)}}$ $IF = \left( {1 + {x^2}} \right)$ Multiplying by (1+x2) to equation (1) we get $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = \cot x$ $\frac{d}{{dx}}\left[ {\left( {1 + {x^2}} \right)y} \right] = \cot x$ $\int {\frac{d}{{dx}}\left[ {\left( {1 + {x^2}} \right)y} \right]} dx = \int {\cot xdx}$ (1+x2)y = log|sinx|+c $y = \frac{{\log \left| {\sin x} \right|}}{{1 + {x^2}}} + \frac{c}{{1 + {x^2}}}$

Question (9)

$x\frac{{dy}}{{dx}} + y - x + xy\cot x = 0\quad \left( {x \ne 0} \right)$

Solution

$x\frac{{dy}}{{dx}} + y\left( {1 + x\cot x} \right) = x$ $\frac{{dy}}{{dx}} + \frac{{y\left( {1 + x\cot x} \right)}}{x} = 1$ $\frac{{dy}}{{dx}} + \left( {\frac{1}{x} + \cot x} \right)y = 1 - - - (1)$ comparing with standard equation we get $p = \frac{1}{x} + \cot x$ $IF = {e^{\int {pdx} }}$ $IF = {e^{\int {\left( {\frac{1}{x} + \cot x} \right)dx} }}$ $IF = {e^{\log x + \log \sin x}}$ $IF = {e^{\log \left( {x\sin x} \right)}}$ IF=xsinx
Multiplying by xsinx $x\sin x\frac{{dy}}{{dx}} + \left( {\frac{1}{x} + \cot x} \right)y \cdot x\sin = x\sin x$ $\frac{d}{{dx}}\left( {x\sin x \cdot y} \right) = x\sin x$ $\int {\frac{d}{{dx}}\left( {x\sin x \cdot y} \right)dx} = \int {x\sin xdx}$ $x\sin x \cdot y = x\int {\sin xdx - \int {\left( {\frac{d}{{dx}}x\int {\sin x \cdot dx} } \right)} } dx$ $x\sin x \cdot y = - x\cos x - \int {1\left( { - \cos x} \right)dx}$ $x\sin x \cdot y = - x\cos x + \sin x + c$ $y = - \cot x + \frac{1}{x} + \frac{c}{{x\sin x}}$ $y = \frac{1}{x} - \cot x + \frac{c}{{x\sin x}}$

Question (10)

$\left( {x + y} \right)\frac{{dy}}{{dx}} = 1$

Solution

$x + y = \frac{{dx}}{{dy}}$ $\frac{{dx}}{{dy}} - x = y - - - (1)$ comparing to standard equation $\frac{{dx}}{{dy}} + px = q\left( y \right)$ p = -1 $IF = {e^{\int {pdy} }}$ $IF = {e^{\int { - 1dy} }}$ $IF = {e^{ - y}}$ Multiplying by IF to equaion (1) we get ${e^{ - y}}\frac{{dx}}{{dy}} - x{e^{ - y}} = y{e^{ - y}}$ $\frac{d}{{dy}}\left( {x{e^{ - y}}} \right) = y{e^{ - y}}$ $\int {\frac{d}{{dy}}\left( {x{e^{ - y}}} \right)dy} = \int {y{e^{ - y}}dy}$ $x{e^{ - y}} = y\int {{e^{ - y}}dy - \int {\left( {\frac{d}{{dy}}y\int {{e^{ - y}}dy} } \right)dy} }$ $x{e^{ - y}} = \frac{{y{e^{ - y}}}}{{ - 1}} - \int {\frac{{{e^{ - y}}}}{{ - 1}}dy}$ $x{e^{ - y}} = - y{e^{ - y}} + \frac{{{e^{ - y}}}}{{ - 1}} + c$ $x = - y - 1 + c{e^y}$ x+y+1=cey

Question (11)

ydx +(x - y2) dy = 0

Solution

ydx=-(x-y2)dy $y\frac{{dx}}{{dy}} = - x + {y^2}$ $y\frac{{dx}}{{dy}} + x = {y^2}$ $\frac{{dx}}{{dy}} + \frac{x}{y} = y - - - (1)$ comparing to standard equation $\frac{{dx}}{{dy}} + px = q\left( y \right)$ p = 1/y $IF = {e^{\int {pdy} }}$ $IF = {e^{\int {\frac{1}{y}dy} }}$ $IF = {e^{\log y}}$ $IF = {y^{\log e}} = y$ Multiply by y to equation (1) we get $y\frac{{dx}}{{dy}} + x = {y^2}$ $\frac{d}{{dy}}\left( {xy} \right) = {y^2}$ $\int {\frac{d}{{dy}}\left( {xy} \right)dy} = \int {{y^2}dy}$ $xy = \frac{{{y^3}}}{3} + c$ $x = \frac{{{y^2}}}{3} + \frac{c}{y}$

Question (12)

$\left( {x + 3{y^2}} \right)\frac{{dy}}{{dx}} = y\quad \left( {y > 0} \right)$

Solution

$\frac{{x + 3{y^2}}}{y} = \frac{{dx}}{{dy}}$ $\frac{x}{y} + 3y = \frac{{dx}}{{dy}}$ $\frac{{dx}}{{dy}} - \frac{x}{y} = 3y - - - (1)$ p = -1/y $IF = {e^{\int {pdy} }}$ $IF = {e^{\int {\frac{{ - 1}}{y}dy} }}$ $IF = {e^{ - \log y}}$ $IF = {y^{ - \log e}} = {y^{ - 1}} = \frac{1}{y}$ Multiply by 1/y to equation (1) we get $\frac{1}{y}\frac{{dx}}{{dy}} - \frac{x}{{{y^2}}} = 3$ $\frac{d}{{dy}}\left( {\frac{x}{y}} \right) = 3$ $\int {\frac{d}{{dy}}\left( {\frac{x}{y}} \right)dy} = 3\int {dy}$ $\frac{x}{y} = 3y + c$
x = 3y2 + cy
For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition :

Question (13)

$\frac{{dy}}{{dx}} + 2y\tan x = \sin x$ y=0 when x = π/3

Solution

comparing with standered equation $\frac{{dy}}{{dx}} + py = q\left( x \right)$ p=2tanx $IF = {e^{\int {pdx} }}$ $IF = {e^{\int {2\tan xdx} }}$ $IF = {e^{2\log \sec x}}$ $IF = {\sec ^{2\log e}}x$ IF = sec2 x
Multiplying by sec2x to equation (1) ${\sec ^2}x\frac{{dy}}{{dx}} + 2y\tan x{\sec ^2}x = \sin x{\sec ^2}x$ $\frac{d}{{dx}}\left( {{{\sec }^2}x \cdot y} \right) = \frac{{\sin x}}{{{{\cos }^2}x}}$ $\frac{d}{{dx}}\left( {{{\sec }^2}x \cdot y} \right) = \tan x\sec x$ $\int {\frac{d}{{dx}}\left( {{{\sec }^2}x} \right)ydx = \int {\sec x\tan xdx} }$ sec2x (y) = secx + c $y = \frac{1}{{\sec x}} + \frac{c}{{{{\sec }^2}x}}$ y= cosx + c cos2x
y = 0 when x = π/3 $y = \cos \frac{\pi }{3} + c{\cos ^2}\frac{\pi }{3}$ $\frac{1}{2} + \frac{1}{4}c = 0$ $\frac{1}{4}c = - \frac{1}{2}$ c=-2
y= cosx - 2cos2x

Question (14)

$\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = \frac{1}{{1 + {x^2}}}$ y= 0 when x=1

Solution

$\frac{{dy}}{{dx}} + \frac{{2x}}{{1 + {x^2}}}y = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} - - - (1)$ comparying to standard form $\frac{{dy}}{{dx}} + py = q\left( x \right)$ $p = \frac{{2x}}{{1 + {x^2}}}$ $IF = {e^{\int {pdx} }}$ $IF = {e^{\int {\frac{{2x}}{{1 + {x^2}}}dx} }}$ $IF = {e^{\log \left( {1 + {x^2}} \right)}}$ $IF = {\left( {1 + {x^2}} \right)^{\log e}}$ IF = (1+x2)
Multiply by (1+x2) to equation (1) $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = \frac{1}{{1 + {x^2}}}$ $\frac{d}{{dx}}\left[ {\left( {1 + {x^2}} \right)y} \right] = \frac{1}{{1 + {x^2}}}$ $\int {\frac{d}{{dx}}\left[ {\left( {1 + {x^2}} \right)y} \right]dx} = \int {\frac{1}{{1 + {x^2}}}dx}$ (1+x2)y = tan-1x + c $y = \frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}} + \frac{c}{{1 + {x^2}}}$ y = 0 when x = 1 $0 = \frac{{{{\tan }^{ - 1}}x}}{{1 + 1}} + \frac{c}{2}$ $\frac{c}{2} = - \frac{{\frac{\pi }{4}}}{2}$ $c = - \frac{\pi }{4}$ $y = \frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}} - \frac{{\frac{\pi }{4}}}{{1 + {x^2}}}$

Question (15)

$\frac{{dy}}{{dx}} - 3y\cot x = \sin 2x$ y = 2 when x=π/2

Solution

compare to standard form we get p= -3cotx $IF = {e^{\int { - 3\cot xdx} }}$ $IF = {e^{ - 3\log \sin x}}$ $IF = {\left( {\sin x} \right)^{ - 3\log e}}$ $IF = {\left( {\sin x} \right)^{ - 3}} = \frac{1}{{{{\sin }^3}x}}$ Multiply by IF to equation (1) wqe get $\frac{1}{{{{\sin }^3}x}}\frac{{dy}}{{dx}} - \frac{{3y\cot x}}{{{{\sin }^3}x}} = \sin 2x\frac{1}{{{{\sin }^3}x}}$ $\left( {\frac{1}{{{{\sin }^3}x}}y} \right) = 2\require{cancel}\cancel{sin x}\cos x \times \frac{1}{{{{\sin}^\cancel{3}}x}}$ $\frac{d}{{dx}}\left( {\frac{1}{{{{\sin }^3}x}}y} \right) = 2\cot x cosec x$ $\int {\frac{d}{{dx}}\left( {\frac{1}{{{{\sin }^3}x}}y} \right)dx} = 2\int {\cot x\cos xdx}$ $\frac{y}{{{{\sin }^3}x}} = - 2 cosecx + c$ y = -2sin2x + c sin3x
y = 2 when x = π/2
2 = -2sin2(π/2) + c sin3π/2
2 = -2 + c
c = 4
y = 4 sin3x-2sin2x

Question (16)

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point

Solution

Curve passing through (0, 0)
Slope of tangent = sum of co-ordinate $\frac{{dy}}{{dx}} = x + y$ $\frac{{dy}}{{dx}} - y = x - - - (1)$ compare to standard form we get p = -1 $IF = {e^{\int {pdx} }}$ $IF = {e^{\int { - dx} }}$ $IF = {e^{ - x}}$ Multiplying by e-x to equation (1) ${e^{ - x}}\frac{{dy}}{{dx}} - y{e^{ - x}} = x{e^{ - x}}$ $\therefore \frac{d}{{dx}}\left( {y{e^{ - x}}} \right) = x{e^{ - x}}$ $\int {\frac{d}{{dx}}\left( {y{e^{ - x}}} \right)dx} = \int {x{e^{ - x}}dx}$ $y{e^{ - x}} = x\int {{e^{ - x}}dx} - \int {\left( {\frac{d}{{dx}}x\int {{e^{ - x}}dx} } \right)dx}$ $y{e^{ - x}} = \frac{{x{e^{ - x}}}}{{ - 1}} - \int {1\frac{{{e^{ - x}}}}{{ - 1}}dx}$ ye-x = -xe-x - e-x + c
y = -x - 1 + cex
It is passing through (0, 0)
∴ 0 = 0 - 1 + ce0
1 = c
y = -x - 1 + ex
x+y+1 = ex

Question (17)

Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5

Solution

Curve passing through (0, 2)
Sum of co-ordinate > slope of tanget by 2
$x + y > \frac{{dy}}{{dx}}\quad \left( {by\;5} \right)$ $x + y = \frac{{dy}}{{dx}} + 5$ $\frac{{dy}}{{dx}} = x + y - 5$ $\frac{{dy}}{{dx}} - y = x - 5 - - - (1)$ comparing to standard form we get p = -1
$I.F. = {e^{\int {pdx} }}$ $I.F. = {e^{\int { - 1dx} }}$ $I.F. = {e^{ - x}}$ multiplying by I.F.to equation (1) we get
${e^{ - x}}\frac{{dy}}{{dx}} - y{e^{ - x}} = \left( {x - 5} \right){e^{ - x}}$ $\frac{d}{{dx}}\left( {{e^x}y} \right) = x{e^{ - x}} - 5{e^{ - x}}$ $\int {\frac{d}{{dx}}\left( {{e^{ - x}}y} \right)dx} = \int {x{e^{ - x}}dx} - 5\int {{e^{ - x}}dx}$ ${e^{ - x}}y = \left[ {x\int {{e^{ - x}}} dx - \int {\left( {\frac{d}{{dx}}x\int {{e^{ - x}}dx} } \right)dx} } \right] - \frac{{5{e^{ - x}}}}{{ - 1}}$ ${e^{ - x}}y = \frac{{x{e^{ - x}}}}{{ - 1}} - \int {\frac{{1{e^{ - x}}}}{{ - 1}}dx} + 5{e^{ - x}}$ ${e^{ - x}}y = - x{e^{ - x}} - {e^{ - x}} + 5{e^{ - x}} + c$ ∴ y = -x - 1+5+ cex
x+y-4= cex
it passes through (0, 2)
0 + 2 -4 = ce0
-2 = c
x + y - 4 = -2ex
x + y = 4 -2ex

Question (18)

The Integrating Factor of the differential equation $x\frac{{dy}}{{dx}} - y = 2{x^2}$ (A) e-x   (B) e-y   (C) 1/x   (D) x

Solution

$x\frac{{dy}}{{dx}} - y = 2{x^2}$ p = 1
$IF = {e^{\int {pdx} }} = {e^{\int { - dx} }}$ $IF = {e^{ - x}}$ Correct option (A)

Question (19)

The Integrating Factor of the differential equation $\left( {1 - {y^2}} \right)\frac{{dx}}{{dy}} + yx = ay$ (-1 < y <) is $\left( A \right)\frac{1}{{{y^2} - 1}}$ $\left( B \right)\frac{1}{{\sqrt {{y^2} - 1} }}$ $\left( C \right)\frac{1}{{1 - {y^2}}}$ $\left( D \right)\frac{1}{{\sqrt {1 - {y^2}} }}$

Solution

$\frac{{dx}}{{dy}} + \frac{y}{{1 - {y^2}}}x = \frac{a}{{1 - {y^2}}}y$ comparing with stanadrd equation $\frac{{dx}}{{dy}} + px = q\left( x \right)$ we get $p = \frac{y}{{1 - {y^2}}}$ $IF = {e^{\int {pdx} }}$ $IF = {e^{\int {\frac{y}{{1 - {y^2}}}dx} }}$ $IF = {e^{ - \frac{1}{2}\int {\frac{{ - 2y}}{{1 - {y^2}}}dx} }}$ $IF = {e^{ - \frac{1}{2}\log \left| {1 - {y^2}} \right|}}$ $IF = {\left( {1 - {y^2}} \right)^{ - \frac{1}{2}\log e}}$ $IF = {\left( {1 - {y^2}} \right)^{ - \frac{1}{2}}}$ $IF = \frac{1}{{\sqrt {1 - {y^2}} }}$ Correct option is (D)