Concise reasoning: - Order = highest order derivative occurring. - Degree = power of the highest order derivative after clearing fractions and making the equation a polynomial in derivatives; if the equation involves non‑algebraic functions of derivatives (sin, cos, integrals, exponentials multiplying derivatives in nonpolynomial way) degree is not defined. Apply these rules to each part: (i) has dy/dx linearly → order 1, degree 1. (ii) highest derivative is d^3y/dx^3 appearing squared → order 3 degree 2. (iii) and (viii),(ix) involve nonpolynomial functions → degree does not exist. For the remaining parts the highest derivative orders are as stated above and appear to first power → degree 1.
(i) order = 1, degree = 1. (ii) order = 3, degree = 2. (iii) order = 2, degree does not exist (non‑polynomial because of sin). (iv) order = 7, degree = 1 (highest derivative assumed 7 and appears linearly). (v) order = 3, degree = 1. (vi) order = 2, degree = 1. (vii) order = 2, degree = 1. (viii) order = 2, degree does not exist (contains cos — non‑polynomial in derivatives). (ix) order = 1, degree does not exist (contains an integral — not algebraic in derivatives). (x) order = 1, degree = 1 (derivative appears to first power).
Direct translation of proportionality statements: rate = constant × indicated quantity (with negative sign for decay). For (iv) continuous 8% gives 0.08A; continuous credit ₹400 adds +400 to dA/dt.
(i) \(\dfrac{dQ}{dt}=-kQ\), \(k>0\). (ii) \(\dfrac{dP}{dt}=kP(500000-P)\). (iii) \(\dfrac{dP}{dT}=k\dfrac{P}{T^2}\). (iv) If A(t) is amount, \(\dfrac{dA}{dt}=0.08A+400\).
Differentiate V: dV/dt = 4π r^2 (dr/dt). Equate to -k·S = -k·4π r^2 and cancel 4π r^2 (r>0) to get dr/dt = -k (constant).
Let V=(4/3)π r^3 and surface area S=4π r^2. dV/dt = -k S = -4π k r^2 ⇒ 4π r^2 dr/dt = -4π k r^2 ⇒ dr/dt = -k.
A straight line has two arbitrary constants; eliminating them by differentiation twice gives the second derivative zero. For lines expressed as x(y) the same idea gives d^2x/dy^2=0.
(i) For non‑vertical lines y=mx+c ⇒ y''=0. (ii) For non‑horizontal lines x=ny+c (view x as function of y) ⇒ d^2x/dy^2=0. (If expressed as y(x) and excluding horizontal lines, the family is still y=mx+c with m≠0 and y''=0.)
Tangent condition: distance from centre to line equals radius; algebraically for y=mx+c tangency gives c^2=r^2(1+m^2). Replace m by y' and c by y-xy' (value of c from y=mx+c) to obtain the DE shown.
For line y=mx+c to be tangent to circle: c^2=r^2(1+m^2). Using m=y' and c=y-xy' gives differential equation: (y-xy')^2 = r^2(1+(y')^2).
General circle centre (a,0) passing through (0,0): (x-a)^2+y^2=a^2 ⇒ x^2+y^2-2ax=0. Differentiate: 2x+2yy'-2a=0 ⇒ a=x+yy'. Substitute a into original: x^2+y^2-2x(x+yy')=0 ⇒ y^2-x^2-2xyy'=0 ⇒ y^2-x^2=2xyy'.
Differential equation: y^2 - x^2 = 2 x y y' .
General parabola with axis parallel to x‑axis: (y-k)^2=4a(x-h). Differentiate: 2(y-k)y'=4a ⇒ y-k=2a/y'. Differentiate again: 2(y')^2+2(y-k)y''=0. Substitute y-k from above: 2(y')^2+2(2a/y')y''=0 ⇒ (y')^3+2a y''=0.
Differential equation: (y')^3 + 2a y'' = 0.
Parabola with axis along y: x^2=4a(y+1). Differentiate: 2x = 4a y' ⇒ a = x/(2y'). Substitute into original: x^2 = 4(x/(2y'))(y+1) ⇒ x y' = 2(y+1) ⇒ x y' - 2y - 2 = 0.
Differential equation: x y' = 2y + 2.
General ellipse: x^2/a^2 + y^2/b^2 =1. Differentiate: x/a^2 + y y'/b^2 =0 ⇒ express 1/b^2 and substitute into original to obtain b^2 = y^2 - x y y'. Since b is constant differentiate b^2: d/dx(y^2 - x y y')=0 ⇒ 2y y' - [y y' + x((y')^2 + y y'')] =0. Simplify to x y y'' + x (y')^2 - y y' =0.
Differential equation: x y y'' + x (y')^2 - y y' = 0.
Differentiate twice: y'' = (1/64)(A e^{x/8} + B e^{-x/8}) = (1/64) y ⇒ 64 y'' - y = 0.
Differential equation: 64 y'' - y = 0 (equivalently y'' - (1/64) y = 0).
Given xy = ae x + be -x + x² ……… (1)
where a & b are aribitrary constant,
differentiate equation (1) twice successively,
because we have two arbitray constant.
From (1), we get xy – x² = ae x + be -x …….. (4)
Substituting equation (4) in (3), we get
∴ x $\frac { d^2y }{ dx^2 }$ + $\frac { 2dy }{ dx }$ – xy + x² – 2 = 0 is the required differential equation.
From (1), we get xy – x² = ae x + be -x …….. (4) Substituting equation (4) in (3), we get ∴ x $\frac { d^2y }{ dx^2 }$ + $\frac { 2dy }{ dx }$ – xy + x² – 2 = 0 is the required differential equation.
Differentiate the proposed solutions and substitute into the given differential equations to verify identity.
(i) y=x^2 ⇒ y'=2x ⇒ x y' = x·2x = 2x^2 = 2y, holds. (ii) y=Ae^x+Be^{-x} ⇒ y''=Ae^x+Be^{-x}=y ⇒ y''-y=0, holds.
Substitute y=e^{mx} and cancel e^{mx} (nonzero) to obtain algebraic equations for m and solve.
(i) Interpreting as y'+y=0 ⇒ for y=e^{mx}: m e^{mx}+e^{mx}=0 ⇒ m+1=0 ⇒ m=-1. (ii) For y=e^{mx}: m^2 - m - 6 = 0 ⇒ (m-3)(m+2)=0 ⇒ m=3 or m=-2.
Separable ODE solved by integration and constant determined from initial point.
Integrate: dy/dx = 1/(4y) ⇒ 4y dy = dx ⇒ ∫4y dy = ∫ dx ⇒ 2 y^2 = x + C. Use (2,5): 2·25 = 2 + C ⇒ 50 = 2 + C ⇒ C = 48. Hence 2 y^2 = x + 48 or x = 2 y^2 - 48.
Given y = e^{-x} + mx + n. Then dy/dx = -e^{-x} + m and d²y/dx² = e^{-x}. Therefore e^x(d²y/dx²) - 1 = e^x e^{-x} - 1 = 1 - 1 = 0. Hence the given function satisfies the differential equation.
e^x(d²y/dx²) - 1 = 0, so y = e^{-x} + mx + n is a solution.
$y=ax+\dfrac bx$ gives $y'=a-\dfrac{b}{x^2}$ and $y''=\dfrac{2b}{x^3}$. Substituting: $x^2y''+xy'-y=x^2\!\left(\dfrac{2b}{x^3}\right)+x\!\left(a-\dfrac{b}{x^2}\right)-\left(ax+\dfrac bx\right)=\dfrac{2b}{x}+ax-\dfrac bx-ax-\dfrac bx=0$. Hence $y=ax+\dfrac bx$ satisfies the equation.
Substituting $y=ax+\dfrac bx$ makes $x^2y''+xy'-y=0$ identically, so it is a solution.
Given y = ae -3x + b …… (1)
Differentiating equation (1) w.r.t ‘x’, we get
Therefore, y = ae -3x + b is a solution of the given differential equation.
Given y = ae -3x + b …… (1) Differentiating equation (1) w.r.t ‘x’, we get Therefore, y = ae -3x + b is a solution of the given differential equation.
Given y² = 2ax + 2a 5/3 ……. (1)
Differentiating equation (1) w.r.t ‘x’ we get
Hence y² = 2ox + 2 a³ is a solution of the differential equation
(y² – 2xy$\frac { dy }{ dx }$)³ = 8(y$\frac { dy }{ dx }$) 5
Differentiating equation (1) w.r.t ‘x’ we get Hence y² = 2ox + 2 a³ is a solution of the differential equation (y² – 2xy$\frac { dy }{ dx }$)³ = 8(y$\frac { dy }{ dx }$) 5
Differentiate: y' = -ab sin(bx), y'' = -ab^2 cos(bx) = -b^2(a cos(bx)) = -b^2 y.
Hence y'' + b^2 y = 0, so y = a cos(bx) satisfies the equation.
Linear ODE: dV/dt + (k/M)V = F/M. Integrating factor e^{(k/M)t}. Solve: V e^{(k/M)t} = (F/k)(e^{(k/M)t}-1) ⇒ V=(F/k)(1-e^{-(k/M)t}). Initial condition V(0)=0 satisfied.
Solution: V(t) = (F/k) (1 - e^{- (k/M) t}).
Given v dv/dx = g - (k/2) v^2. Put A = k/2. Then v/(g - A v^2) dv = dx. Integrate: \int v/(g-A v^2) dv = -\frac{1}{2A}\ln(g-A v^2)=x+C. With v(0)=0: -\frac{1}{2A}\ln g = C. Hence \ln(g-A v^2)=\ln g -2Ax, so g-A v^2 = g e^{-2Ax}. Thus v^2 = (g/A)(1-e^{-2Ax}). Substituting A=k/2 gives v^2=(2g/k)(1-e^{-kx}). Taking positive root (downward speed): v=\sqrt{\dfrac{2g}{k}(1-e^{-kx})}.
v = \sqrt{\dfrac{2g}{k}\bigl(1-e^{-kx}\bigr)}
We interpret dy/dx = (y-x)/(x+2). Rewrite: dy/dx - y/(x+2) = -x/(x+2). IF = e^{-\int dx/(x+2)}=1/(x+2). Multiply: d/dx(y/(x+2)) = -x/(x+2)^2. Integrate: y/(x+2) = \int -\frac{x}{(x+2)^2}dx = -\ln(x+2) -\frac{2}{x+2} + C. Hence y=-(x+2)\ln(x+2) -2 + C(x+2). Use (1,0): 0= -3\ln3 -2 +3C => C=(2+3\ln3)/3. Substitute for final y.
y = -(x+2)\ln(x+2) -2 + C(x+2) with C=\tfrac{2+3\ln 3}{3}; so y=-(x+2)\ln(x+2)-2 + (\tfrac{2+3\ln 3}{3})(x+2).
(i) $\frac { dy }{ dx }$ = $\sqrt { 1-y^2 }{ 1-x^2 }$
(ii) ydx + (1 + x²) tan -1 x dy = 0
(iii) sin $\frac { dy }{ dx }$ = a, y (0) = 1
(iv) $\frac { dy }{ dx }$e x+y + x³, e y
(v) (e y + 1) cos x dx + e y sin x dy = 0
(vi) (ydx – xdy) cot ($\frac { x }{ y }$) = ny² dx
(vii) $\frac { dy }{ dx }$ – x$\sqrt { 25-x^2 }$ = 0
(viii) x cos y dy = e x (x log x + 1) dx
(xi) tan y $\frac { dy }{ dx }$ = cos (x + y) + cos (x – y)
(x) $\frac { dy }{ dx }$ = tan² (x + y)v
(i) $\frac { dy }{ dx }$ = $\sqrt { 1-y^2 }{ 1-x^2 }$
The equation can be written as
$\frac { dy }{ \sqrt {1-y^2} }$ = $\frac { dx }{ \sqrt {1-x^2} }$
Taking Integration on both sides, we get
∫ $\frac { dy }{ \sqrt {1-y^2} }$ = ∫ $\frac { dx }{ \sqrt {1-x^2} }$
sin -1 y = sin -1 x + C
(ii) ydx + (1 + x²) tan -1 x dy = 0
ydx = – (1 + x²) tan -1 x dy
Take t = tan -1 x
dt = $\frac { 1 }{ 1+x^2 }$ dx
The equation can be written as
$\frac { dx }{ (1+x^2)tan^{-1}x }$ = –$\frac { dy }{ y }$
$\frac { dt }{ t }$ = –$\frac { dy }{ y }$
Taking Integration on both sides, we get
∫ $\frac { dt }{ t }$ = ∫ $\frac { dy }{ y }$
log t = – log y + log C
log (tan -1 x) = – log y + log C
log (y(tan -1 x)) + log y = log C
y tan -1 x = C
(iii) sin $\frac { dy }{ dx }$ = a, y (0) = 1
sin $\frac { dy }{ dx }$ = a
sin $\frac { dy }{ dx }$ = sin -1 (a)
The equation can be written as
dy = sin -1 (a) dx
Taking integration on both sides, we get
∫ dy = ∫ sin -1 (a) dx
y = sin -1 a ∫ dx
y = sin -1 (a) x + C ……… (1)
Initial condition:
Since y (0) = 1, we get
y = sin -1 (a) x + C
1 = sin -1 (a) (0) + C
0 + C = 1
C = 1
Equation (1) ⇒ y = sin -1 (a) x + 1
y – 1 = sin -1 (a) x
$\frac { y-1 }{ x }$ = sin -1 (a)
sin($\frac { y-1 }{ x }$) = a
(iv) $\frac { dy }{ dx }$e x+y + x³, e y
$\frac { dy }{ dx }$e x+y + x³, (e y )
= e y [e x + x³]
$\frac { dy }{ e^y }$ = dx(e x + x³)
The equation can be written as
$\frac { dy }{ e^y }$ = (e x + x³) dx
Taking integration on both sides, we get
∫ e -y dy = ∫ (e x + x³) dx
$\frac { e^y }{ -1 }$ = e x + $\frac { x^4 }{ 4 }$ + C
[Where – C = C, Which is also constant].
∴ e x + e -y + $\frac { x^4 }{4 }$ = -C = C
∴ e x + e -y + $\frac { x^4 }{4 }$ = C
(v) (e y + 1) cos x dx + e y sin x dy = 0
(e y + 1) cos x dx + e y sin x dy = 0
e y sin x dy = – (e y + 1) cos x dx
log (e y + 1) = – log sin x + log c
log [(e y + 1) + log sin x = log c
log (e y +1) sin x] = log c
(e y + 1) sin x = c
(vi) (ydx – xdy) cot ($\frac { x }{ y }$) = ny² dx
equation can be written as
Substituting these values in equation (1), we get
dt cot t = ndx
cot t dt = ndx
Taking integration on both sides, we get
∫ cot t dt = n∫ dx
log (sin t) = n x + c
sin t = e nx+c
∴ sin($\frac { x }{ y }$) = e nx+c [∵ t = $\frac { x }{ y }$ ]
(vii) $\frac { dy }{ dx }$ – x$\sqrt { 25-x^2 }$ = 0
The equation can be written as
$\frac { dy }{ dx }$ – x$\sqrt { 25-x^2 }$ …….. (1)
Take 25 – x² = t
-2x dx = dt
x dx = –$\frac { dt }{ 2 }$
Substituting these values in equation (1), we get
dy = x$\sqrt { 25-x^2 }$ dx
dy = -√t $\frac { dt }{ 2 }$
Taking integration on both sides, we get
∫ dy = –$\frac { dt }{ 2 }$ ∫ t $\frac { 1 }{ 2 }$ dt
(viii) x cos y dy = e x (x log x + 1) dx
The equation can be written as
Substituting in (1), we get
sin y = e y log x + C
(ix) tan y $\frac { dy }{ dx }$ = cos (x + y) + cos (x – y)
The equation can be written as
tan y $\frac { dy }{ dx }$ = cos (x + y) + cos (x – y)
[W.K.T cos (A + B) + cos (A – B) = 2 cos A cos B
Here A = x, B = y]
∴ tan y$\frac { dy }{ dx }$ = 2 cos x cos y
$\frac { tany }{ cosy }$ dy = 2 cos x dx
Taking integration on both sides, we get
∫ $\frac { tany }{ cosy }$ dy = 2 ∫ cos x dx
2 ∫ tan y sec y dy = 2 ∫ cos x dx
sec y = 2 sin x + C
(x) $\frac { dy }{ dx }$ = tan² (x + y)
2 ∫ tan y sec y dy = 2 ∫ cos x dx sec y = 2 sin x + C (x) $\frac { dy }{ dx }$ = tan² (x + y)
[x + y cos($\frac { y }{ x }$)] dx = x cos ($\frac { y }{ x }$) dyv
The given equation can be written as
On integration we obtain
which gives the required solution.
The given equation can be written as On integration we obtain which gives the required solution.
The given equation can be written as
The given equation can be written as
Given equation is (y² – 2xy) dx = (x² – 2xy) dy
y² – 2xy = (x² – 2xy) $\frac { dy }{ dx }$
∴ The equation can written as
log (3v² – 3v) = – 3 log x + log C
log (3v² – 3v) = – log x³ + log C
= log c – log x³
log (3v² – 3v) = – 3 log x + log C log (3v² – 3v) = – log x³ + log C = log c – log x³
The given differential equation may be
Given that y = 0 when x = 1
0 + 3(1) e° = c
3 = c
∴ y + 3xe y/x = 3 is a required solution.
0 + 3(1) e° = c 3 = c ∴ y + 3xe y/x = 3 is a required solution.
The given differential equation is of the form
The given differential equation is of the form
cos x $\frac { dy }{ dx }$ + y sin x = 1v
The given differential equation can be written as
This is the form $\frac { dy }{ dx }$ + py = Q
where P = tan x
Q = sec x
Thus, the given differential equation is linear.
I.F = e ∫ pdx = e ∫ tan x dx = e log (sec x) = sec x
So, the required solution is given by
[y × I.F] = ∫ [Q × IF] dx + c
y × sec x = ∫ sec x × sec x dx + c
y sec x = ∫ sec² x dx + c
y sec x = tan x + c
= sin x + c cos x
y = sin x + c cos x is the required solution.
y sec x = tan x + c = sin x + c cos x y = sin x + c cos x is the required solution.
This is linear of the form $\dfrac{dy}{dx}+Py=Q$ with $P=\dfrac1x$ and $Q=\sin x$. The integrating factor is $e^{\int\frac1x\,dx}=e^{\log x}=x$. Multiplying through, $\dfrac{d}{dx}(xy)=x\sin x$, so $xy=\displaystyle\int x\sin x\,dx=-x\cos x+\sin x+c$.
$xy=\sin x-x\cos x+c$.
The given differential equation may be written as
y dx = -(2x – 10y³) dy
So, the required solution is
x × I.F = ∫ (Q × I.F)dy + c
xy² = ∫ 10 y² × y² dy + c
= ∫ 10 y 4 dy + c
= $\frac { 10y^5 }{ 5 }$ + c = 2y 5 + c
xy² = 2y 5 + c is a required solution.
= ∫ 10 y 4 dy + c = $\frac { 10y^5 }{ 5 }$ + c = 2y 5 + c xy² = 2y 5 + c is a required solution.
Thus, the given differential equation is Linear
Thus, the given differential equation is Linear
So, the solution of the equation is given by
xy + tan -1 y = c
Which is the required solution.
So, the solution of the equation is given by xy + tan -1 y = c Which is the required solution.
The given linear differential equation is of the form
$\frac { dy }{ dx }$ + py = Q
The given linear differential equation is of the form $\frac { dy }{ dx }$ + py = Q
The equation can be written as
The equation can be written as
The given differential equation may be written as
Thus, the given differential equation is linear.
I.F = e ∫pdx = e ∫$\frac { 1 }{ x }$ dx = e log x = x
So, the solution of the given differential equation is given by
y × I.F = ∫(Q × I.F)dx + c
yx = ∫ log x x dx + c
yx = ∫ x log x dx + c
u = log x ∫dv = ∫x dx
Multiply by 4
4yx = 2x² log x – x² + 4c
4xy = 2x² log x – x² + 4c is a required solution.
Multiply by 4 4yx = 2x² log x – x² + 4c 4xy = 2x² log x – x² + 4c is a required solution.
The given differential equation can be written as
So, its solution is given by
y × I.F = ∫(Q × I.F) dx + c
yx³ = ∫ $\frac { 1 }{ x^2 }$ + x³ dx + c
= ∫x dx + c
y x³ = $\frac { x^2 }{ 2 }$ + c
2yx³ = x² + c
Given that y = 2 when x = 1
2 (2) (1)³ = 1 + c
4 – 1 = c
c = 3
∴ 2yx³ = x² + 3 is a required solution.
4 – 1 = c c = 3 ∴ 2yx³ = x² + 3 is a required solution.
Model N' = kN => N(t)=N_0 e^{kt}. Given N(5)=3N_0 => e^{5k}=3 => k=(1/5)\ln 3. Then N(10)=N_0 e^{10k}=N_0 e^{2\ln3}=N_0 3^2 =9N_0.
N(10)=N_0 \cdot 9
P' = kP => P(t)=P_0 e^{kt}. Given P(40)=400000=300000 e^{40k} => e^{40k}=4/3 => k=\tfrac{1}{40}\ln(4/3). Thus P(t)=300000 e^{(\ln(4/3)/40) t} =300000 (4/3)^{t/40}.
P(t)=300000 \cdot \bigl(\tfrac{4}{3}\bigr)^{t/40} =300000\,e^{(\ln(4/3)/40) t}
With E=0: L di/dt + R i = 0 => di/dt = -(R/L) i. Solve: i(t)=C e^{-\tfrac{R}{L}t}. Using initial current i(0)=i_0 gives i(t)=i_0 e^{-R t / L}.
i(t)=i(0) e^{-\tfrac{R}{L}t}
dv/dt = −v, v(0)=10. Solve: v = 10 e^{-t}. At t=2: v(2)=10 e^{-2} ≈1.353 m/s.
v(2)=10e^{-2} m/s ≈ 1.353 m/s
Continuous compounding: A = P e^{rt} with P=10000, r=0.05, t=1.5. So A=10000 e^{0.075} ≈10000×1.07788≈₹10,778.8.
A = 10000 e^{0.05×1.5} = 10000 e^{0.075} ≈ ₹10,778.8
N' = −kN ⇒ N(t)=N_0 e^{-kt}. Given N(100)=0.9N_0 ⇒ e^{-100k}=0.9. Then N(1000)=N_0 e^{-1000k}=(e^{-100k})^{10}=0.9^{10}≈0.348678 ⇒ about 34.87%.
Remaining fraction after 1000 yr = 0.9^{10} ≈ 0.348678 => 34.8678% remain
Newton's law: T(t)=25+75 e^{-kt}. From T(10)=80 ⇒ 25+75 e^{-10k}=80 ⇒ e^{-10k}=11/15, so k=-(1/10)ln(11/15).
(i) T(20)=25+75 e^{-20k}=25+75(e^{-10k})^2=25+75(11/15)^2=25+75·121/225=25+40.333…=65.33°C (approx).
(ii) For T=40: 40=25+75 e^{-kt} ⇒ e^{-kt}=1/5 ⇒ t=-(1/k)ln(0.2). Using k=-(1/10)ln(11/15) gives t≈51.9 minutes.
Model: T(t)=70+110 e^{-kt}. From T(10)=160 ⇒ 70+110 e^{-10k}=160 ⇒ e^{-10k}=9/11, so k=(1/10)ln(11/9).
(i) T(15)=70+110 e^{-15k}=70+110(e^{-10k})^{1.5}=70+110(9/11)^{1.5}≈151.39°F.
(ii) Solve 70+110 e^{-kt}=140 ⇒ e^{-kt}=70/110=0.63636 ⇒ t≈22.5 min. For 130°F: e^{-kt}=60/110=0.54545 ⇒ t≈30.2 min. Thus she should drink between about 10:22.5 A.M. and 10:30.2 A.M. (≈10:22:30 to ≈10:30:12).
Let room temperature be r and T(t)=r+(100−r)e^{-kt}. Put u=e^{-5k}. From T(5)=80 ⇒ r+(100−r)u=80 and T(10)=65 ⇒ r+(100−r)u^2=65.
Subtracting gives (100−r)u(1−u)=15, and from the first equation (100−r)(1−u)=20. Dividing gives u=15/20=0.75. Then (100−r)(1−0.75)=20 ⇒ (100−r)·0.25=20 ⇒ 100−r=80 ⇒ r=20°C.
dS/dt = in − out = 3×2 − 3(S/50) = 6 − (3/50)S. Solve linear: dS/dt + (3/50)S = 6. Integrating factor e^{3t/50}. Solution S=100 + C e^{-3t/50}. With S(0)=0 ⇒ C=−100. Thus S(t)=100(1−e^{-3t/50}).
S(t)=100(1−e^{-3t/50}) grams
If the highest derivative present is of order 4 and derivatives appear to first power, the order is 4 and the degree (power of highest derivative when equation is a polynomial in derivatives) is 1.
Order = 4, Degree = 1
y = A cos x + B ⇒ y' = −A sin x, y'' = −A cos x. From y'' + y = B (constant). Differentiating to eliminate B gives y''' + y' = 0, which is satisfied by the family.
Hence the ODE with no arbitrary constants is y''' + y' = 0.
The displayed form suggests a first-order equation linear in dy/dx (no fractional powers). Hence order =1 and degree =1.
Option (3): order 1, degree 1
A general circle (x−h)^2+(y−k)^2=a^2 contains three arbitrary constants h, k and a. The differential equation representing this family therefore has order 3.
Differentiate twice: y'' = A e^x + B e^{-x} = y. Hence y'' − y = 0, which contains no arbitrary constants.
Solution. Separating variables, dy/y = dx/x. Hence log y = log x + C, so y = kx.
Answer: Option (3): y = kx.
(A) straight lines (B) circles (C) parabola (D) ellipsev
(c) parabola
Hint:
2 log (3 + y) = log x + log k
log (3 + y)² = log kx
Remove log, we get
(3 + y)² = kx is a solution of the differential equation which is a Parabola.
parabola
Standard linear homogeneous equation: integrating factor e^{∫ p dx} gives y e^{∫ p dx} = C ⇒ y = C e^{-∫ p dx}.
Option (2): y = C e^{-∫ p(x) dx}
(A) $\frac { x }{ e^x }$ + y = 0 (B) $\frac { e^x }{ x }$ – y = 0 (C) λe x (D) e xv
(b) $\frac { e^x }{ x }$ – y = 0
Hint:
Given differential equation is
$\frac { e^x }{ x }$ – y = 0
μ(x)=e^{∫P dx}=x ⇒ ∫P dx = ln x ⇒ P(x)=1/x.
Option (3): 1/x
(A) $2$ (B) $3$ (C) $1$ (D) $4$v
(c) $1$
The bracketed series is the expansion of $e^{\,dy/dx}$, so $y=x\,e^{\,dy/dx}$. Hence $\dfrac yx=e^{\,dy/dx}$, giving $\dfrac{dy}{dx}=\log\dfrac yx$. The highest-order derivative $\dfrac{dy}{dx}$ appears to the first power, so the degree is $1$.
$1$
(A) $p
(c) $p>q$
The highest-order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $p=2$; it occurs to the first power, so the degree is $q=1$. Therefore $p>q$.
$p>q$
(A) y + sin -1 x = c (B) x + sin -1 y = 0 (C) y² + 2sin -1 x = c (D) x² + 2sin -1 y = 0v
(a) y + sin -1 x = c
Hint:
The given equation is $\frac { dy }{ dx }$ = –$\frac { 1 }{ \sqrt{1-x^2} }$
The equation can be written as
dy = $\frac { 1 }{ \sqrt{1-x^2} }$
Integrating on both sides, we get
∫dy = -∫$\frac { dx }{ \sqrt{1-x^2} }$
y = -sin -1 (x) + C
y + sin -1 (x) = C
∴ y + sin -1 (x) = C is a solution of the given differential equation.
y + sin -1 x = c
Separate variables: dy/y = 2x dx. Integrate: ln|y| = x^2 + C => y = C e^{x^2}.
From log(dy/dx)=x+y => dy/dx = e^{x+y}=e^x e^y. Separate: e^{-y} dy = e^x dx. Integrate: -e^{-y}=e^x + C => e^x + e^{-y}=C.
Write dy/dx - (1/x)y = -2. IF μ=e^{∫(-1/x)dx}=1/x. Then (y/x)' = -2/x. Integrate: y/x = -2 ln|x| + C => y = x(C - 2 ln|x|).
For a homogeneous DE dy/dx = φ(y/x), put v = y/x => y = vx, dy/dx = v + x dv/dx. Then v + x dv/dx = φ(v) => x dv/dx = φ(v)-v. Separate and integrate to obtain an implicit relation which can be expressed as x = k·F(v) i.e. x = k φ(y/x) (form (1)).
If μ = sin x is an integrating factor then μ = e^{∫P dx} => ∫P dx = ln(sin x) => P = d/dx(ln sin x) = cos x / sin x = cot x.
A differential equation of order n has a general solution containing n arbitrary constants; order n+1 has n+1 arbitrary constants.
A particular (specific) solution has no arbitrary constants, so for any order the particular solution contains 0 arbitrary constants.
IF μ = e^{∫P dx} with P=1/(x+1) gives μ = e^{ln|x+1|}=|x+1|. Take μ=x+1.
dP/dt = kP => dP/P = k dt => ln P = kt + C => P = C e^{kt}.
dP/dt = -kP => dP/P = -k dt => P = C e^{-k t}.
For the family to represent a circle the quadratic terms in x and y coming from integration must balance appropriately; the standard result gives a = -2. (Answer supplied with low confidence due to ambiguous statement.)
Integrate: y = ∫3x^2 dx = x^3 + C. Use point (-1,1): 1 = (-1)^3 + C => C=2. So y = x^3 + 2.