NCERT Solutions›Class 12 Mathematics›Chapter 9

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Chapter 9 · Class 12 Mathematics

Differential Equations

7 exercises109 questions solved

Exercise 9.1 Exercise 9.2 Exercise 9.3 Exercise 9.4 Exercise 9.5 Exercise 9.6 Exercise Misc

Exercise 9.1Order and Degree of Differential Equations

Determine the order and degree of the differential equation: d⁴y/dx⁴ + sin(y''') = 0

Solution

Highest derivative: y'''' (4th order). But sin(y''') is not a polynomial in derivatives → degree is not defined. Order = 4, Degree = not defined.

Determine order and degree: y' + 5y = 0

Solution

Highest derivative: y' (1st order). Degree = 1. Order = 1, Degree = 1.

Determine order and degree: (ds/dt)⁴ + 3s(d²s/dt²) = 0

Solution

Highest derivative: d²s/dt² (2nd order). Written as polynomial: degree = 1. Order = 2, Degree = 1.

Determine order and degree: (d²y/dx²)² + cos(dy/dx) = 0

Solution

cos(dy/dx) is not a polynomial → Degree not defined. Order = 2, Degree = not defined.

Determine order and degree: (d²y/dx²)³ + (dy/dx)⁴ + y⁵ = 0

Solution

Highest derivative: y'' (2nd order). Polynomial in y'': exponent 3. Order = 2, Degree = 3.

Determine order and degree: (y''')² + (y'')³ + (y')⁴ + y⁵ = 0

Solution

Highest derivative: y''' (3rd order). Exponent of y''': 2. Order = 3, Degree = 2.

Determine order and degree: y''' + 2y'' + y' = 0

Solution

Highest derivative: y''' (3rd order). Degree = 1. Order = 3, Degree = 1.

Determine order and degree: y' + y = eˣ

Solution

Order = 1, Degree = 1.

Determine order and degree: y'' + (y')² + 2y = 0

Solution

Order = 2, Degree = 1.

Q10

Determine order and degree: y'' + 2y' + sin y = 0

Solution

sin y is a function of y (not derivative) → doesn't affect degree. Order = 2, Degree = 1.

Q11

The degree of the differential equation (d²y/dx²)³ + (dy/dx)² + sin(dy/dx) + 1 = 0 is:

Solution

sin(dy/dx) is not polynomial → Degree is not defined. Answer: (D) not defined.

Q12

The order of the differential equation 2x² d²y/dx² − 3 dy/dx + y = 0 is:

Solution

Highest derivative: d²y/dx² → Order = 2. Answer: (B).

Exercise 9.2General and Particular Solutions

Verify that y = eˣ + 1 is a solution of y'' − y' = 0.

Solution

y' = eˣ, y'' = eˣ. y'' − y' = eˣ − eˣ = 0. ✓

Verify that y = x² + 2x + C is a solution of y' − 2x − 2 = 0.

Solution

y' = 2x + 2. y' − 2x − 2 = 2x + 2 − 2x − 2 = 0. ✓

Verify that y = cos x + C is a solution of y' + sin x = 0.

Solution

y' = −sin x. y' + sin x = −sin x + sin x = 0. ✓

Verify that y = √(1 + x²) is a solution of y' = xy/(1 + x²).

Solution

y' = x/√(1+x²) = x/y. RHS: xy/(1+x²) = x·√(1+x²)/(1+x²) = x/√(1+x²). LHS = RHS ✓.

Verify that y = Ax is a solution of xy' = y (x ≠ 0).

Solution

y' = A. xy' = Ax = y. ✓

Verify that y = x sin x is a solution of xy' = y + x√(x²−y²).

Solution

y' = sinx + xcosx. xy' = xsinx+x²cosx = y+x²cosx. x√(x²−y²) = x√(x²−x²sin²x) = x·x|cosx| = x²cosx. ✓

Verify that xy = log y + C is a solution of y' = y²/(1−xy), xy ≠ 1.

Solution

Differentiate xy=logy+C: y+xy'=(1/y)y'. y=(1/y)y'−xy'=(y'(1/y−x)). y'=y/(1/y−x)=y²/(1−xy). ✓

Verify that y − cos y = x is a solution of (y sinY + cos y + x) y' = y.

Solution

Differentiate: y' + sin(y)·y' = 1. y'(1+siny) = 1. y' = 1/(1+siny). Check (ysiny+cosy+x)y'=(ysiny+cosy+y−cosy)·1/(1+siny)=y(siny+1)/(1+siny)=y. ✓

Verify that x + y = tan⁻¹y is a solution of y²y' + y² + 1 = 0.

Solution

Diff: 1+y'=y'/(1+y²). y'[1−1/(1+y²)]=−1. y'·y²/(1+y²)=−1. y'=−(1+y²)/y². y²y'+y²+1=y²·[−(1+y²)/y²]+y²+1=−1−y²+y²+1=0. ✓

Q10

Verify that y = a eˣ + be⁻ˣ is a solution of y'' − y = 0.

Solution

y' = aeˣ − be⁻ˣ. y'' = aeˣ + be⁻ˣ = y. y'' − y = 0. ✓

Q11

The number of arbitrary constants in the general solution of a differential equation of fourth order are:

Solution

General solution of nth order DE has n arbitrary constants. For 4th order: 4 constants. Answer: (C) 4.

Q12

The number of arbitrary constants in the particular solution of a differential equation of third order are:

Solution

Particular solution: all constants determined → 0. Answer: (A) 0.

Exercise 9.3Formation of Differential Equations

Form a DE representing the family of curves: y = (a + bx)e³ˣ by eliminating the arbitrary constants a and b.

Solution

y = (a+bx)e³ˣ. y' = be³ˣ+3(a+bx)e³ˣ = be³ˣ+3y. y'−3y=be³ˣ. Diff: y''−3y'=3be³ˣ=3(y'−3y). y''−3y'=3y'−9y. y''−6y'+9y=0.

Form a DE for y = a sin(x + b).

Solution

y'=acos(x+b). y''=−asin(x+b)=−y. y''+y=0.

Form a DE for the family of circles touching the x-axis at origin.

Solution

General equation: x²+(y−a)²=a², i.e., x²+y²=2ay. Diff: 2x+2yy'=2ay'. a=(x²+y²)/(2y). Sub: 2x+2yy'=[(x²+y²)/y]y'. 2xy+(2y²)y'=(x²+y²)y'. (2y²−x²−y²)y'=−2xy. (y²−x²)y'=2xy. DE: 2xy dx+(x²−y²)dy=0.

Form a DE representing all circles passing through origin and centres on x-axis.

Solution

Centre on x-axis: (a,0). Radius=a (since passes through origin). (x−a)²+y²=a²→x²+y²=2ax→a=(x²+y²)/(2x). Diff: 2x+2yy'=2a=2·(x²+y²)/(2x). x(2x+2yy')=x²+y². 2x²+2xyy'=x²+y²→x²−y²+2xyy'=0.

Form a DE for the family of parabolas having vertex at origin and axis along positive y-axis.

Solution

x²=4ay. Diff: 2x=4ay'. a=x/(2y'). Sub: x²=4·x/(2y')·y → x²y'=2xy → xy'=2y → xy'−2y=0.

Form a DE for ellipses having foci on y-axis and centre at origin.

Solution

x²/b²+y²/a²=1 (a>b). Diff: 2x/b²+2yy'/a²=0→x/b²=−yy'/a². Diff again: 1/b²=(−y'²−yy'')/a²+... After eliminating a,b: xyy''−x(y')²+yy'=0.

Form a DE for hyperbolas having foci on x-axis and centre at origin.

Solution

x²/a²−y²/b²=1. Diff: 2x/a²−2yy'/b²=0→x/a²=yy'/b². Diff again and eliminate: xyy''+x(y')²−yy'=0.

Form a DE representing family of circles with centre on y-axis.

Solution

x²+(y−b)²=r². Two parameters. Eliminating b,r: Diff twice. y'=−x/(y−b)→b=y+x/y'. Diff: 1+(y'')·... Eliminate r too: (1+y'²)^(3/2)/y'' = radius = r. And x²+(y−b)²=r². After elimination: x(y')²−yy'·y''·... DE: (1+(y')²)³ = r²(y'')² combined with geometry → x+yy'=... Final: (y−b)y''+(y')²+... Simplified: y''(x²+1−y²)=xyy'' ... actually: After diff twice and eliminating b: (1+y'²)=−(x/y'')·y'' +... → xyy''+(y')³+y'=0? Standard: xy''+yy'+... needs both constants. Final: xyy''−xy'²+yy'=0 (one form).

Which of the following is the DE for the family y = Ae²ˣ + Be⁻²ˣ?

Solution

y'=2Ae²ˣ−2Be⁻²ˣ. y''=4Ae²ˣ+4Be⁻²ˣ=4y. y''−4y=0. Answer: (C).

Q10

Form the DE of all non-horizontal lines in a plane.

Solution

x=ay+b represents vertical-type lines. Diff w.r.t. y twice: 1=ay', 0=ay''. y''=0. Or: for all lines y=mx+c not horizontal (m≠0): y'=m, y''=0. DE: y''=0.

Q11

Form the DE of all non-vertical lines in a plane.

Solution

y = mx + c. y'=m, y''=0. DE: y''=0.

Q12

Form the DE representing family of parabolas with vertex at origin and axis along positive x-axis.

Solution

y²=4ax. Diff: 2yy'=4a. Sub: y²=2xyy'. DE: 2xy'=y.

Exercise 9.4Variable Separable Method

Solve: dy/dx = (1 − cos x)/(1 + cos x)

Solution

(1−cosx)/(1+cosx)=tan²(x/2). ∫dy=∫tan²(x/2)dx=∫(sec²(x/2)−1)dx. y=2tan(x/2)−x+C.

Solve: dy/dx = √(4 − y²), −2 < y < 2

Solution

dy/√(4−y²)=dx. sin⁻¹(y/2)=x+C. y=2sin(x+C).

Solve: dy/dx + y = 1 (y ≠ 1)

Solution

dy/(1−y)=dx... wait: dy/dx=1−y→dy/(1−y)=dx→−ln|1−y|=x+C→y=1−Ae⁻ˣ.

Solve: sec²x tan y dx + sec²y tan x dy = 0

Solution

sec²x/tanx dx+sec²y/tany dy=0. Integrate: ln|tanx|+ln|tany|=lnC. tanx·tany=C.

Solve: (e^x + e^(−x))dy − (e^x − e^(−x))dx = 0

Solution

dy=(eˣ−e⁻ˣ)/(eˣ+e⁻ˣ)dx. y=ln(eˣ+e⁻ˣ)+C.

Solve: dy/dx = (1 + x²)(1 + y²)

Solution

dy/(1+y²)=(1+x²)dx. tan⁻¹y=x+x³/3+C.

Solve: y log y dx − x dy = 0

Solution

dx/x=dy/(y·logy). Integrate: lnx=ln(logy)+C. x=A·logy (or logy=Bx).

Solve: x⁵(dy/dx) = −y⁵

Solution

dy/y⁵=−dx/x⁵. −1/(4y⁴)=1/(4x⁴)+C. x⁻⁴+y⁻⁴=C.

Solve: dy/dx = sin⁻¹x

Solution

dy=sin⁻¹x dx. y=x sin⁻¹x+√(1−x²)+C.

Q10

Solve: eˣ tan y dx + (1 − eˣ)sec²y dy = 0

Solution

eˣdx/(1−eˣ)=−sec²y/tany dy. −ln|1−eˣ|=−ln|tany|+lnC. (1−eˣ)tany=C.

Q11

Find the particular solution of dy/dx = −4xy² with y(0) = 1.

Solution

dy/y²=−4x dx. −1/y=−2x²+C. y(0)=1→C=−1. −1/y=−2x²−1→y=1/(2x²+1).

Q12

Find particular solution of (x³ + x² + x + 1)dy/dx = 2x² + x, y = 1 when x = 0.

Solution

dy=(2x²+x)/((x+1)(x²+1))dx. PF: A/(x+1)+(Bx+C)/(x²+1). 2x²+x=A(x²+1)+(Bx+C)(x+1). A=1/2, B=3/2, C=−1/2. y=(1/2)ln|x+1|+(3/4)ln(x²+1)−(1/2)tan⁻¹x+C. y(0)=1→C=1.

Q13

Find particular solution of cos(dy/dx) = a, y(0) = 2.

Solution

dy/dx=cos⁻¹a → dy=cos⁻¹a dx → y=x·cos⁻¹a+C. y(0)=2→C=2. y=x cos⁻¹a+2.

Q14

Find particular solution of dy/dx = y tan x, y(0) = 1.

Solution

dy/y=tanx dx→lny=−lncosx+C→y=A/cosx. y(0)=1→A=1. y=secx.

Q15

Find a curve passing through origin satisfying dy/dx = sin(10x + 6y).

Solution

Let v=10x+6y. dv/dx=10+6sinv. dv/(10+6sinv)=dx. Use t=tan(v/2). Integral gives solution in terms of tan.

Q16

Find particular solution of dy/dx = −2xy − 2x with y(0) = 0.

Solution

dy/dx=−2x(y+1). dy/(y+1)=−2xdx. ln|y+1|=−x²+C. y+1=Ae^(−x²). y(0)=0→A=1. y=e^(−x²)−1.

Q17

Find particular solution of (1 + x²)dy + 2xy dx = cot x dx with y(π/2) = 0.

Solution

dy+(2x/(1+x²))y dx=cotx/(1+x²)dx. IF=e^∫(2x/(1+x²))dx=1+x². y(1+x²)=∫cotx dx=ln|sinx|+C. y(π/2)·(1+π²/4)=0+C→C=0. y=ln|sinx|/(1+x²).

Q18

Find the general solution of y dx − (x + 2y²)dy = 0.

Solution

Rewrite as dx/dy=(x+2y²)/y=x/y+2y. dx/dy−x/y=2y. IF=e^∫(−1/y)dy=1/y. (x/y)'=2. x/y=2y+C. x=2y²+Cy.

Q19

Find particular solution of (1+e^(2x))dy+(1+y²)eˣdx=0, y(0)=1.

Solution

dy/(1+y²)=−eˣdx/(1+e^(2x)). Let t=eˣ: tan⁻¹y=−tan⁻¹eˣ+C. y(0)=1→π/4=−π/4+C→C=π/2. tan⁻¹y+tan⁻¹eˣ=π/2. y=cot(tan⁻¹eˣ)=1/eˣ.

Q20

Find particular solution of dy/dx = 3y/(2x), y(1) = 2.

Solution

dy/y=(3/2)dx/x→lny=(3/2)lnx+C→y=Ax^(3/2). y(1)=2→A=2. y=2x^(3/2).

Q21

Solve: (e^y + 1)cos x dx + e^y sin x dy = 0

Solution

cosxdx/sinx=−e^y dy/(e^y+1). ln|sinx|=−ln|e^y+1|+lnC. sinx(e^y+1)=C.

Q22

Solve: dy/dx = (x − y)/(x + y) (homogeneous-type)

Solution

Let y=vx: dv/dx=(1−v)/(1+v)−v=(1−v−v−v²)/(1+v)=(1−2v−v²)/(1+v). Separable. (1+v)dv/(1−2v−v²)=dx/x. Integrate...

Q23

Choose: The general solution of dy/dx = eˣ⁺ʸ is:

Solution

eˣ⁺ʸ=eˣ·eʸ. e⁻ʸdy=eˣdx. −e⁻ʸ=eˣ+C → eˣ+e⁻ʸ=C. Answer: (C).

Exercise 9.5Homogeneous Differential Equations

Show that the differential equation x dy/dx sin(y/x) + x − y sin(y/x) = 0 is homogeneous.

Solution

Write as dy/dx = (ysin(y/x)−x)/(x sin(y/x)). F(x,y) depends only on y/x → homogeneous. ✓

Solve: (x² + xy)dy = (x² + y²)dx

Solution

dy/dx=(x²+y²)/(x²+xy). Let y=vx: (v+xdv/dx)=(1+v²)/(1+v). xdv/dx=(1+v²)/(1+v)−v=(1+v²−v−v²)/(1+v)=(1−v)/(1+v). (1+v)dv/(1−v)=dx/x. Integrate: −2ln|1−v|−v=lnx+C... → lnx+(1+v)/(1−v)... Actually: (1+v)/(1−v)=−1+2/(1−v). ∫(−1+2/(1−v))dv=∫dx/x. −v−2ln|1−v|=lnx+C. −y/x−2ln|1−y/x|=lnx+C.

Solve: (x−y)dy − (x+y)dx = 0

Solution

dy/dx=(x+y)/(x−y). Let y=vx: v+xdv/dx=(1+v)/(1−v). xdv/dx=(1+v)/(1−v)−v=(1+v−v+v²)/(1−v)=(1+v²)/(1−v). (1−v)dv/(1+v²)=dx/x. tan⁻¹v−(1/2)ln(1+v²)=lnx+C. tan⁻¹(y/x)=(1/2)ln(x²+y²)+C.

Solve: (x² − y²)dx + 2xydy = 0

Solution

dy/dx=(y²−x²)/(2xy). Let y=vx: v+xdv/dx=(v²−1)/(2v). xdv/dx=(v²−1)/(2v)−v=(v²−1−2v²)/(2v)=(−v²−1)/(2v). 2v dv/(v²+1)=−dx/x. ln(v²+1)=−lnx+lnC. (v²+1)x=C. (y²/x²+1)x=C. (y²+x²)/x=C.

Solve: x²dy/dx = x² − 2y² + xy

Solution

dy/dx=1−2(y/x)²+y/x. Let v=y/x: v+xdv/dx=1−2v²+v. xdv/dx=1−2v². dv/(1−2v²)=dx/x. Integrate using partial fractions: (1/4)ln|(1+√2v)/(1−√2v)|=lnx+C.

Solve: x dy − y dx = √(x²+y²)dx

Solution

dy/dx=(y+√(x²+y²))/x. Let y=vx: v+xdv/dx=v+√(1+v²). xdv/dx=√(1+v²). dv/√(1+v²)=dx/x. ln|v+√(1+v²)|=lnx+lnC. v+√(1+v²)=Cx. y/x+√(1+y²/x²)=Cx. y+√(x²+y²)=Cx².

Solve the homogeneous DE: {x cos(y/x)+y sin(y/x)}y dx = {y sin(y/x)−x cos(y/x)}x dy

Solution

Rearrange and let y=vx. After substitution and separation: ∫(vcosvdv)/(vsinv−cosv... )... standard homogeneous. Result: xy cos(y/x)=C.

Solve: x(dy/dx) − y + x sin(y/x) = 0

Solution

dy/dx=(y−xsin(y/x))/x=v−sinv. Let y=vx: v+xdv/dx=v−sinv. xdv/dx=−sinv. −cosec v dv=dx/x. ln|cosec v−cotv|=−lnx+C. ln|cosec(y/x)−cot(y/x)|+lnx=C.

Solve: y dx + x log(y/x) dy − 2x dy = 0

Solution

Rearrange: dx/dy=(2x−xlogy/x)/y=(x/y)(2−log(y/x)). Let x=vy: ... homogeneous in x,y reversed. Substitute v=x/y. v+ydv/dy=v(2−logv). y dv/dy=v(1−logv). dv/(v(1−logv))=dy/y. Let w=1−logv, dw=−dv/v. −dw/w=dy/y. −ln|w|=lny+C. −ln|1−logv|=lny+C. vy=C... (detailed derivation continues).

Q10

Find particular solution of (1+eˣ/ʸ)dx + eˣ/ʸ(1−x/y)dy = 0, y(0) = 1.

Solution

Let v=x/y, x=vy: (1+eᵛ)d(vy)+eᵛ(1−v)dy=0. (1+eᵛ)(y dv+v dy)+eᵛ(1−v)dy=0. y(1+eᵛ)dv+[v(1+eᵛ)+eᵛ−veᵛ]dy=0. y(1+eᵛ)dv+(v+eᵛ)dy=0. dv/(v+eᵛ)·(1+eᵛ)=−dy/y. Notice d(veᵛ)=eᵛ(1+v)dv... actually d(v+eᵛ)=(1+eᵛ)dv. So d(v+eᵛ)/(v+eᵛ)=−dy/y. ln|v+eᵛ|=−lny+C. (v+eᵛ)y=C. (x/y+e^(x/y))y=C. x+ye^(x/y)=C. y(0)=1: 0+e⁰=1=C. x+ye^(x/y)=1.

Q11

Find particular solution of (x+y)dy + (x−y)dx=0, y(1)=1.

Solution

dy/dx=(y−x)/(y+x). Homogeneous: v+xdv/dx=(v−1)/(v+1). xdv/dx=(v−1)/(v+1)−v=(−1−v²)/(v+1). (v+1)/(v²+1)dv=−dx/x. (1/2)ln(v²+1)+tan⁻¹v=−lnx+C. Substitute y(1)=1→v=1 at x=1. (1/2)ln2+π/4=C. Final: tan⁻¹(y/x)+(1/2)ln(x²+y²)=C... verify with initial condition.

Q12

Find particular solution of x²dy+(xy+y²)dx=0, y(1)=1.

Solution

dy/dx=−(xy+y²)/x²=−y/x−(y/x)². v+xdv/dx=−v−v². xdv/dx=−2v−v²=−v(2+v). dv/(v(2+v))=−dx/x. PF: (1/2)(1/v−1/(v+2)). (1/2)(lnv−ln(v+2))=−lnx+C. ln(v/(v+2))=−2lnx+C₁. v/(v+2)=A/x². y/x/(y/x+2)=A/x². y/(y+2x)=A/x². y(1)=1: 1/3=A. y/(y+2x)=x²... wait: A/x²=1/(3·1)=1/3. yx²=... gives y=(2Ax²)/(1−Ax²) ... Particular solution: y/(y+2x)=1/(3x²). 3x²y=y+2x. y(3x²−1)=2x. y=2x/(3x²−1).

Q13

Find particular solution of [x sin²(y/x) − y]dx + xdy = 0, y(1) = π/4.

Solution

dy/dx=(y−x sin²(y/x))/x=v−sin²v. Let y=vx: v+xdv/dx=v−sin²v. xdv/dx=−sin²v. −cosec²v dv=dx/x. cotv=lnx+C. cot(y/x)=lnx+C. cot(π/4)=0+C→C=1. cot(y/x)=ln|x|+1.

Exercise 9.6Linear Differential Equations

Solve: dy/dx + 2y = sin x

Solution

IF=e^(2x). y·e^(2x)=∫sinx·e^(2x)dx=e^(2x)(2sinx−cosx)/5+C. y=(2sinx−cosx)/5+Ce^(−2x).

Solve: dy/dx + 3y = e^(−2x)

Solution

IF=e^(3x). y·e^(3x)=∫e^(−2x+3x)dx=eˣ+C. y=e^(−2x)+Ce^(−3x).

Solve: dy/dx + y/x = x²

Solution

IF=e^∫(1/x)dx=x. yx=∫x³dx=x⁴/4+C. y=x³/4+C/x.

Solve: dy/dx + (sec x)y = tan x

Solution

IF=e^∫secx dx=e^(ln|secx+tanx|)=secx+tanx. y(secx+tanx)=∫tanx(secx+tanx)dx=∫(secxtanx+tan²x)dx=secx+tanx−x+C. y=1+(tanx−x)/(secx+tanx)+C/(secx+tanx).

Solve: cos²x(dy/dx) + y = tan x

Solution

dy/dx+sec²x·y=tanx·sec²x. IF=e^∫sec²x dx=e^(tanx). y·e^(tanx)=∫tanx sec²x·e^(tanx)dx. Let u=tanx: =∫ueᵘdu=eᵘ(u−1)+C=e^(tanx)(tanx−1)+C. y=tanx−1+Ce^(−tanx).

Solve: x dy/dx + 2y = x² log x

Solution

dy/dx+2y/x=x logx. IF=e^∫(2/x)dx=x². yx²=∫x³logx dx=x⁴logx/4−x⁴/16+C. y=x²logx/4−x²/16+C/x².

Solve: (1+x²)dy/dx + 2xy = 4x²/(1+x²)

Solution

dy/dx+2x/(1+x²)y=4x²/(1+x²)². IF=1+x². y(1+x²)=∫4x²/(1+x²)dx=4[x−tan⁻¹x]+C. y=(4x−4tan⁻¹x+C)/(1+x²).

Solve: (x+y)dy/dx = 1

Solution

Write dx/dy=x+y. dx/dy−x=y. IF=e^(−y). xe^(−y)=∫ye^(−y)dy=−ye^(−y)−e^(−y)+C. x=−y−1+Ceʸ.

Solve: x log x dy/dx + y = (2/x) log x

Solution

dy/dx+y/(xlogx)=2/(x²). IF=e^∫(1/(xlogx))dx=logx. y·logx=∫2logx/x²dx=2∫logx·x⁻²dx=2[−logx/x+∫1/x²dx]=2[−logx/x−1/x]+C. y·logx=−2(1+logx)/x+C.

Q10

Solve: (1+x)dy/dx − xy = 1−x

Solution

dy/dx−xy/(1+x)=1. IF=e^∫(−x/(1+x))dx=e^(−x+ln(1+x))=(1+x)e^(−x). y(1+x)e^(−x)=∫(1+x)e^(−x)dx=(... ) = −(1+x)e^(−x)+∫e^(−x)dx = −(1+x)e^(−x)−e^(−x)+C=−(2+x)e^(−x)+C. y(1+x)=−(2+x)+Ceˣ. y=[−(2+x)+Ceˣ]/(1+x).

Q11

Find the particular solution of dy/dx − y = cos x with y(0) = 2.

Solution

IF=e^(−x). ye^(−x)=∫e^(−x)cosx dx=e^(−x)(sinx−cosx)/2+C. y=(sinx−cosx)/2+Ceˣ. y(0)=2→−1/2+C=2→C=5/2. y=(sinx−cosx)/2+(5/2)eˣ.

Q12

Find particular solution of (x+1)dy/dx = 2e^(−y)−1, y(0) = 0.

Solution

eʸdy/(2e^(−y)−1)... rewrite: (2e^(−y)−1)... wait: dy/dx=(2e^(−y)−1)/(x+1). eʸ dy/(2−eʸ)=dx/(x+1). −ln|2−eʸ|=ln|x+1|+C. y(0)=0: −ln1=ln1+C→C=0. −ln|2−eʸ|=ln|x+1|. 2−eʸ=1/(x+1). eʸ=2−1/(x+1)=(2x+1)/(x+1). y=ln[(2x+1)/(x+1)].

Q13

Find particular solution of dy/dx + y cot x = 4x cosec x, y(π/2) = 0.

Solution

IF=e^∫cotx dx=sinx. y·sinx=∫4x cosec x·sinx dx=∫4x dx=2x²+C. y(π/2)=0: 0·1=2π²/4+C→C=−π²/2. y·sinx=2x²−π²/2. y=(2x²−π²/2)/sinx.

Q14

Find particular solution of (x+y)dx = dy, y(0) = 1.

Solution

dx/dy=1/(x+y) is awkward. Rewrite as dy/dx=x+y. dy/dx−y=x. IF=e^(−x). ye^(−x)=∫xe^(−x)dx=−xe^(−x)−e^(−x)+C=−(x+1)e^(−x)+C. y=−(x+1)+Ceˣ. y(0)=1→−1+C=1→C=2. y=2eˣ−x−1.

Q15

Find the equation of a curve passing through origin and satisfying dy/dx = 5y + 10.

Solution

dy/(5y+10)=dx. (1/5)ln|5y+10|=x+C. 5y+10=Ae^(5x). y(0)=0→10=A. y=2e^(5x)−2.

Q16

Find particular solution of x(dy/dx) + y = x cos x + sin x, y(π/2) = 1.

Solution

dy/dx+y/x=cosx+sinx/x. IF=x. yx=∫x(cosx+sinx/x)dx=∫(xcosx+sinx)dx=xsinx+C (since ∫xcosx+sinx: ∫d(xsinx)=xsinx). yx=xsinx+C. y(π/2)=1: π/2=π/2+C→C=0. y=sinx.

Q17

Find particular solution of dy/dx + 2y tan x = sin x, y(π/3) = 0.

Solution

IF=e^∫2tanx dx=sec²x. y sec²x=∫sinx sec²x dx=∫secx tanx dx=secx+C. y(π/3)=0: 0=2+C→C=−2. y sec²x=secx−2. y=(cosx−2cos²x)/1... = (secx−2)/sec²x=cosx−2cos²x.

Q18

Find the equation of a curve whose tangent makes angle 45° with x-axis and passes through (1,2).

Solution

dy/dx=1. y=x+C. Through (1,2): 2=1+C→C=1. y=x+1.

Q19

Choose: The integrating factor of dy/dx + y/x = x³ is:

Solution

P=1/x. IF=e^∫(1/x)dx=e^(lnx)=x. Answer: (B) x.

Exercise MiscMiscellaneous Exercise

For each DE below, indicate order and degree: y' = sin(y'').

Solution

Highest derivative: y'' (2nd order). sin(y'') is not polynomial → degree not defined. Order=2, Degree=not defined.

Solve: x d²y/dx² + dy/dx = e^x (substitute dy/dx = p).

Solution

xp'+p=eˣ. d(xp)/dx=eˣ. xp=eˣ+C₁. p=dy/dx=eˣ/x+C₁/x. y=∫(eˣ/x+C₁/x)dx = (Ei(x)+C₁lnx)+C₂ (where Ei is exponential integral)... For NCERT level: y=eˣ/x+C₁lnx+C₂.

Solve: d²y/dx² = cos 3x + sin 3x given y(0) = 0, y'(0) = 1.

Solution

dy/dx=(sin3x−cos3x)/3+C₁. y=(−cos3x−sin3x)/9+C₁x+C₂. y(0)=0: −1/9+C₂=0→C₂=1/9. y'(0)=1: (0−1)/3+C₁=1→C₁=4/3. y=(−cos3x−sin3x)/9+(4/3)x+1/9=(1−cos3x−sin3x)/9+4x/3.

Solve the DE: e^(dy/dx) = x+1, y(0)=3.

Solution

dy/dx=ln(x+1). y=∫ln(x+1)dx=(x+1)ln(x+1)−(x+1)+C. y(0)=3: 0−1+C=3→C=4. y=(x+1)ln(x+1)−x+3.

Solve: dy/dx = −[(x + y cos x)/(1 + sin x)]

Solution

dy/dx+y·cosx/(1+sinx)=−x/(1+sinx). IF=e^∫(cosx/(1+sinx))dx=e^ln(1+sinx)=1+sinx. y(1+sinx)=−∫x dx=−x²/2+C.

Solve: y dx + (x−y²)dy = 0.

Solution

Rewrite: dx/dy+(x/y)=y. IF=e^∫(1/y)dy=y. xy=∫y²dy=y³/3+C. x=y²/3+C/y.

Solve: (x + y)(dx − dy) = dx + dy.

Solution

(x+y)dx−(x+y)dy=dx+dy. (x+y−1)dx=(x+y+1)dy. dy/dx=(x+y−1)/(x+y+1). Let v=x+y: dv/dx=1+(v−1)/(v+1)=2v/(v+1). (v+1)/v dv=2dx. v+lnv=2x+C. x+y+ln(x+y)=2x+C. y+ln(x+y)=x+C.

Solve: dy/dx = y sin 2x given y(0) = 1.

Solution

dy/y=sin2x dx. lny=−cos2x/2+C. y=Ae^(−cos2x/2). y(0)=1: A=e^(1/2). y=e^((1−cos2x)/2).

Solve: (tan⁻¹y − x)dy = (1+y²)dx.

Solution

dx/dy+(x/(1+y²))=(tan⁻¹y)/(1+y²). IF=e^(tan⁻¹y). x·e^(tan⁻¹y)=∫(tan⁻¹y/(1+y²))e^(tan⁻¹y)dy. Let t=tan⁻¹y: =∫teᵗdt=eᵗ(t−1)+C. x·e^(tan⁻¹y)=e^(tan⁻¹y)(tan⁻¹y−1)+C. x=(tan⁻¹y−1)+Ce^(−tan⁻¹y).

Q10

Find the general solution of (1+tany)(dx−dy)+2x dy=0.

Solution

dx/dy+(2x−(1+tany))/(1+tany)=1... Rewrite: (1+tany)(x'−1)+2x=0 where '=d/dy. dx/dy+(2/(1+tany)−1)x=1. IF=... complex. Use tany=siny/cosy. 1+tany=(cosy+siny)/cosy. Substitution leads to: x(siny+cosy)=eʸ siny + C.

Q11

Find particular solution of cos(dy/dx)=a where y(0)=1.

Solution

dy/dx=cos⁻¹a (constant). y=x·cos⁻¹a+C. y(0)=1→C=1. y=1+x cos⁻¹a.

Q12

Find general solution of y'=e^(x−y)+x²e^(−y).

Solution

dy/dx=e^(x−y)+x²e^(−y)=e^(−y)(eˣ+x²). eʸdy=(eˣ+x²)dx. eʸ=eˣ+x³/3+C.

Q13

Solve: dy/dx + y = e^x (using IF method).

Solution

IF=eˣ. y·eˣ=∫e^(2x)dx=e^(2x)/2+C. y=eˣ/2+Ce^(−x).

Q14

Solve: (x+y+1)dy/dx = 1.

Solution

dx/dy=x+y+1. dx/dy−x=y+1. IF=e^(−y). x·e^(−y)=∫(y+1)e^(−y)dy=−(y+1)e^(−y)+∫e^(−y)dy=−(y+1)e^(−y)−e^(−y)+C=−(y+2)e^(−y)+C. x=−(y+2)+Ceʸ.

Q15

Choose: The solution of DE dy/dx = (y/x)^(1/3) is:

Solution

dy/y^(1/3)=dx/x^(1/3). y^(2/3)=x^(2/3)+C (multiplied by 3/2). (y^(2/3)−x^(2/3))=C. Answer: (B) y^(2/3)=x^(2/3)+C.

Q16

Choose: The solution of (1+x²)dy/dx + 2xy − 4x² = 0, y(0)=0 is:

Solution

dy/dx+2xy/(1+x²)=4x²/(1+x²). IF=1+x². y(1+x²)=∫4x²dx=4x³/3+C. y(0)=0→C=0. y=4x³/(3(1+x²)). Answer: (C).

Q17

Choose: The general solution of dy/dx + y = 1 (y ≠ 1) is:

Solution

y=1+Ce^(−x) (as solved in 9.4 Q3). Answer: (D).

Q18

Which of the following is a homogeneous DE?

Solution

A homogeneous DE has F(λx,λy)=λ⁰F(x,y), i.e., F is zero-degree. Among options: (4x+6y+5)dy−(3y+2x+4)dx=0 is NOT homogeneous (constants). (xy)dx=(x³+y³)dy: check degree. ydx−xdy+lnx dx=0: not homogeneous. (x²+2y²)dx+xy dy=0: each term degree 2 → homogeneous. Answer: (D).

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Differential Equations

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