By Newton?s law of cooling w.k.t T = t₂ + (t₁-t₂) e^-kt, given t₁=100?, t₂=25?
when t=1 –> T(1) = 25 + 75 e^-k = 75? –> 50/75 = 2/3 = e^-k
–> 3/2=e^k taking log k=log(1.5)=0.4.
to find T when t=3 minute using value of k we get
T = 25 + 75e^-0.4*3 = 47.5???e^-1.2=0.3.

\(\frac{dy}{dx} – \frac{y}{x} = 1\) is of form \(\frac{dy}{dx}\) + Py = Q where P & Q is a function of x only
given DE is linear DE in y here \(P=\frac{-1}{x}, Q=1, e^{\int P \,dx} = e^{\int \frac{-1}{x} \,dx} = e^{logx^{-1}} = \frac{1}{x}\)
Linear DE solution is given by \(ye^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c \rightarrow y \frac{1}{x} = \int 1 \frac{1}{x} \,dx + c\)
\(\frac{y}{x}\) = log?|x|+c, given y(1)=1–>y=1 when x=1 i.e c=1 refore its particular solution is given by y = x log|x| + x.

\(\frac{dy}{dx} = \frac{y-x+1}{y+x+5}\) –> x=X+h, y=Y+k
h+k+5=0 & k-h+1=0 solving we get h=-2, k=-3
\(\frac{dY}{dX} = \frac{Y-X}{Y+X}\), put Y=vX we get an new equation
\(V + X\frac{dV}{dX} = \frac{v-1}{v+1} \rightarrow X\frac{dV}{dX} = \frac{-(1+v^2)}{1+v}\)
or
\(-\int \frac{1+v}{1+v^2} dv = \int \frac{1}{X} dX = \int \frac{1}{1+v^2} + \frac{v}{1+v^2} \,dv\)
-tan^-1?v-0.5 log(1+v²)=log x +c
\(-tan^{-1} \frac{?Y}{X} – log\sqrt{X^2+Y^2} = c\)
\(-tan{-1}?? \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\) is solution.

By Newton?s law of cooling w.k.t T=t₂+(t₁-t₂) e^-kt, given t₁=72?, t₂=44?
At t=half an hour = 30mts T=61?, finding k using given values i.e
61=44+28e^-k30 –> \(\frac{17}{28}\) = e^-k30 or \(\frac{28}{17}\) = e^k30 taking log, log \(\frac{28}{17}\) = 30k –> k=0.0166
to find T when t = 30mts + 30mts = 60mts
T = 44 + 28e^-(0.0166)30 = 54.4?.

Given: \(\frac{dx}{dt}= \frac{1}{x} \)
dt=xdx
Integrating, \(?dt=?_0^{98}xdx\)
\(t= (\frac{x^2}{2})_0^{98}=\frac{98}{2}=49 s\)

sec² x tan?y dx + sec² y tan?x dy=0
dividing throughout by tan y.tan x we get
\(\frac{sec^2 x}{tan?x} \,dx + \frac{sec^2 y}{tan?y} \,dy=0\)??separating variable
now integrating we get \(\int \frac{sec^2 x}{tan?x} \,dx + \int \frac{sec^2 y}{tan?y} \,dy=c\)
substituting tan x = t & tan y=p?sec² x dx=dt & sec² y dy=dp
\( \rightarrow \int \frac{1}{t} \,dt + \int \frac{1}{p} \,dp=c\)
log t + log p = c –>log(tan x)+log(tan y) = c = log k?.since it is an unknown constant
log(tan x .tan y) = log k
(tan x tan y) = k is solution.

r=a(1+sin ?) taking log we get log r=log a+log(1+sin ?)?differentiating
w.r.t ? we have, \( \frac{1}{r} \frac{dr}{d?} = \frac{cos??}{1+sin??}\) since given equation is polar slope of tangent is given by \( \frac{1}{r} \frac{dr}{d?}\) and perpendicular line has a slope of \(-r\frac{d?}{dr} \, replacing\, \frac{1}{r} \frac{dr}{d?} \,by\, -r\frac{d?}{dr}\)
we have \(-r\frac{d?}{dr} = \frac{cos??}{1+sin??}\) ?..separating variables and integrating it
\(\int \frac{dr}{r} + \int \frac{1+sin??}{cos??} \,d? = c \rightarrow log r + \int sec?? \,d? + \int tan?? \, d? = c\)
log r + log(sec?? + tan??) + log(sec??) = c = log k
\(\rightarrow log(r(sec?? + tan??) (sec??)) = log k \rightarrow r\left(\frac{1}{cos??} + \frac{sin??}{cos??}\right) \frac{1}{cos??} = k\)
\(\frac{r(1+sin??)}{cos^2 ?} = \frac{r(1+sin??)}{1-sin^2 ?} \rightarrow r = k(1-sin ?)\) is required orthogonal trajectory.

Order of a differential equation is given by highest order derivative appearing in differential equation. Hence for given equation, y”+y’-x3y=sinx, order is 2.

\(\frac{dy}{dx} -y \,tan?x = \frac{sin?x cos^2?x}{y^2}\) multiplying by
\(y^2 \rightarrow y^2 \frac{dy}{dx} – y^3 tan?x = sin?x \,cos^2?x \)
put \(t = y^3\rightarrow3y^2 \frac{dy}{dx} = \frac{dt}{dx} \,or\, y^2 \frac{dy}{dx} = \frac{1}{3} \frac{dt}{dx}\)
substituting \(\frac{dt}{dx}-3t \,tan?x = 3sin?x \,cos^2?x \)
this equation is linear in t i.e it is of form \(\frac{dt}{dx} + Pt = Q, e^{\int P \,dx} = e^{\int -3tan?x \,dx}\)
\(e^{-3 log?\, sec?x} = cos^3 \,x\) its solution is \(te^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c\)
t cos³ x=?3sin?x cos²x cos³ x dx + c = ?3sin?x cos⁵?x dx+c, put v=cos x
dv=-sin x dx i.e \(\int 3sin?x \,cos^5\, ?x \,dx = \int -3v^5 \,dx = \frac{-v^6}{2} = \frac{-cos^6 x}{2}\) hence its solution becomes \(t cos^3 x = \frac{-cos^6 x}{2} + c \rightarrow y^3 cos^3 \,x = \frac{-cos^6 x}{2} + c.\)

Let y?=p and hence given equation is in Lagrange equation form
i.e 2y=4xp+log p ?..(1) differentiating both sides of equation
2dy=4xdp+4pdx+\(\frac{dp}{p}\) and dy=pdx
–> 2pdx=4xdp+4pdx+\(\frac{dp}{p}\) –> -2pdx=4pdx+\(\frac{dp}{p}\)
-2p\(\frac{dx}{dp} = 4x + \frac{1}{p} \rightarrow \frac{dx}{dp} + \frac{2}{p} x = \frac{-1}{2p^2}\) (p?0)??.this is a linear D.E for function x(p)
I.F is \(e^{\int \frac{2}{p} \,dp} = e^{log?p^2} = p^2\) and solution is x(p) \(p^2 = \int p^2 *\frac{-1}{2p^2} \,dp + c\)
\(x(p) = \frac{-1}{2p} + \frac{c}{p^2}\) substituting back in (1) we get \(2y=4p\left(\frac{-1}{2p} + \frac{c}{p^2}\right) + log p\)
\(y(p) = \frac{2c}{p} – 1 + \frac{log?p}{2} \).