The given function is y mx +\(\frac{7}{m}\), where m is an arbitrary constant ....(1)
Differentiating both sides of equation (1) with respect to x, we get y' = m.
Substituting the values of y' and y in the given differential equation
we get xy'\(\frac{1}{y'}\)-y = xm +\(\frac{7}{m}\)- mx -\(\frac{7}{m}\) = 0
Therefore, the given function is a solution of the differential equation xy' + 7\(\frac{1}{y'}\) - y = 0

Given that x² + y² = r², r∈R      ...(1)
The given equation contains exactly one arbitrary constant.
So, we have to differentiate the given equation once. Differentiate (1) with respect to x, we get
2x +2y\(\frac{dy}{dx}\) = 0  which implies \(\frac{dy}{dx}\) = \(-\frac{x}{y}\)
Thus, x² + y² = r² satisfies the differential equation \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).
Hence, x² + y² = r² is a solution of the differential equation  \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).

dy + (xy − cos x)dx = 0 is a first order differential equation with degree 1 
since the equation can be rewritten as 
\(\frac{dy}{dx}\) + xy - cos x = 0

The given differential equation is \(3\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 3 }{ 2 } }\)Squaring both sides, we get 
\(9{ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 3 }\)
In this equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 2.
Therefore, the given differential equation is of order 2 and degree 2.

In the given differential equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 1.
Therefore, the given differential equation is of order 2.
The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

Here, the highest order derivative is \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } \) whose power is 3.
Therefore, the given differential equation is of order 4 and degree 3.

In this equation, the highest order derivative is \(\\ \\ \\ \frac { dy }{ dx } \) whose power is 1.
Therefore, the given differential equation is of order 1 and degree 1.

The given equation can be written as \(\frac { dx }{ dy } =\frac { \left( \frac { x }{ y } -1 \right) { 2e }^{ x/y } }{ 1+2{ e }^{ x/y } } =g\left( \frac { x }{ y } \right) ..(1)\)
The appearance of \(\frac{x}{y}\) in equation (1), suggests that the appropriate substitution is x = vy.
Put x = vy . Then, we have \(y\frac { dv }{ dy } =-\frac { 2{ e }^{ v }+v }{ 1+2{ e }^{ v } } \)
By separating the variables, we have \(-\frac { 1+2{ e }^{ v } }{ v+2{ e }^{ v } } dv=-\frac { dy }{ y } \)
On integration, we obtain
log |2e^v + v| = −log |y| + log |C| or log |2ye^v+vy| = log |C| or 2ye^v+ vy = ±C.
Replace v by \(\frac{x}{y}\) to get, 2ye^x/y+x = k, where k =土C, Which gives the required solution.

The given equation is rewritten as \(\frac { dy }{ dx } =\frac { { y }^{ 2 } }{ xy-{ x }^{ 2 } } \)
This is a homogeneous differential equation 
Put y = vx . Then, we have \(x\frac { dv }{ dx } =\frac { v }{ v-1 } \)
By separating the variables, \(\frac { v-1 }{ v } dv=\frac { dx }{ x } .\)
Integrating, we obtain v − log |v| = log |x| + log |C| or v = log |vxC|.
Replacing v by \(\frac{y}{x}\), we get, \(\frac{y}{x}\) = log |Cy| = e^y/x or y = ke^y/x (how!) which is the required solution.

The given differential equation is homogeneous (verify).
Now, we rewrite the given equation in differential form \(\frac { dy }{ dx } =\frac { y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x } \)
Since the initial value of x is 1, we consider x > 0 and take x =\(\sqrt { { x }^{ 2 } } \)
We have \(\frac { dy }{ dx } =\frac { y }{ x } +\sqrt { 1+{ \left( \frac { y }{ x } \right) }^{ 2 } } \)
Let y = vx. Then, \(v+x\frac { dv }{ dx } =v+\sqrt { 1+{ v }^{ 2 } } \), which becomes \(x\frac { dv }{ dx } =\sqrt { 1+{ v }^{ 2 } } \)
By separating variables, we have \(\frac { dv }{ \sqrt { { v }^{ 2 }+1 } } =\frac { dx }{ x } \)
Upon integration, we get \(|v+\sqrt { { v }^{ 2 }+1 } |=log|x|+log|C|\ or\ v+\sqrt { { v }^{ 2 }+1 } =xC\)
Now, we replace v by \(\frac{y}{x}\), we get \(\frac { y }{ x } +\sqrt { \frac { { y }^{ 2 } }{ { x }^{ 2 } } +1 } =Cx\quad (or)\quad y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ Cx }^{ 2 }\) gives the general solution of the given differential equation 
To determine the value of C, we use the condition that y = 0 when x = 1. So, we get C = 1.
Thus \(y=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ x }^{ 2 }\)is the particular solution of the given differential equation.

The given function is y = a cos(log x) + bsin (log x)   ...(1)
where a, b are two arbitrary constants. In order to eliminate the two arbitrary constants, we have to differentiate the given function two times successively
Differentiating equation (1) with respect to x , we get 
y' = -a sin (log x).\(\frac{1}{x}\)+ b cos (log x).\(\frac{1}{x}\Rightarrow\)xy' = -a sin(log x)+b cos (log x).
Again differentiating this with respect to x, we get
xy" + y' = -a cos(log x).\(\frac{1}{x}-b\) sin (log x).\(\frac{1}{x}\Rightarrow\)x²y" + xy'+ y = 0
Therefore, y = a cos(log x) + bsin (log x) is a solution of the given differential equation.

The given function is y = 2(x²−1) + \(Ce^{x^2}\), where C is an arbitrary constant    ... (1)
Differentiating both sides of equation (1) with respect to x, we get \(\frac { dy }{ dx } =4x-2x{ Ce }^{ -x2 }\)
Substituting the values of \(\frac { dy }{ dx } \) and y in the given differential equation, we get
\(\frac { dy }{ dx } \) + 2xy - 4x³ = 4x - 2xCe^-x2 + 2x[2(x²-1)+Ce^-x2]-4x³ = 0
Therefore, the given function is a solution of the differential equation \(\frac { dy }{ dx } \)+2xy-4x³ = 0

Let T be the temperature of the body at any time t and with time 0 taken to be 8 p.m. By Newton’s law of cooling \(\frac { dT }{ dt } =k(T-50)or\frac { dT }{ T-50 } =dt\).
Integrating on both sides, we get log |50 −T| = kt + logC or 50 −T = Ce^kt.
When t = 0, T = 70, and so C = −20
When t = 2,T = 60, we have −10 = −20 e^k2.
Thus, \(k=\frac { 1 }{ 2 } log\left( \frac { 1 }{ 2 } \right) \)
Hence, the solution is 50-T = -20e^{\(\frac{1}{2}\)tlog\((\frac{1}{2})\)} or T = 50 + 20\((\frac{1}{2})^\frac{t}{2}\)
Now, we would like to find the value of t, for which T(t) = 98.6 , and t = 2\(\left( \frac { log\left( \frac { 48.6 }{ 20 } \right) }{ log\left( \frac { 1 }{ 2 } \right) } \right) \approx -2.56\)
It appears that the person was murdered at about 5.30 p.m.

Let A be the mass of the isotope remaining after t years, and let −k be the constant of proportionality, where k > 0. Then the rate of decomposition is modeled by \(\frac{da}{dt}=-kA,\) where the minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables,we get\(\frac{da}{dt}=-kdt\).
Integrating on both sides, we get log |A| = −kt + log |C| or A = Ce−^kt.
Given that the initial mass is 200mg. That is, A = 200 when t = 0 and thus, C = 200.
Thus, we get A = − 200e^-kt.
Also, A =150when t = 2 and therefore, k = \(\frac{1}{2}log(\frac{4}{3})\)
Hence, A(t) = 200e^{\(\frac{1}{2}log(\frac{4}{3})\)} is the mass of isotope remaining after t years.
The half-life t_h is the time corresponding to A = 100 mg
Thus, \({ t }_{ k }=\frac { 2log\left( \frac { 1 }{ 2 } \right) }{ log\left( \frac { 3 }{ 4 } \right) } \).

The given equation can be written as \(\frac { dy }{ dx } =\frac { 2x+3y }{ x-y } \)
This is a homogeneous equation.
Let y = vx . Then we have \(v+x\frac { dv }{ dx } =\frac { 2+3v }{ 1-v } \)
Thus, \(x\frac { dv }{ dx } =\frac { 2+2v+{ v }^{ 2 } }{ 1-v } or\frac { 1-v }{ { (1+v) }^{ 2 }+1 } dv=\frac { dx }{ x } or\frac { 1 }{ 2 } \left[ \frac { 2v+2 }{ { v }^{ 2 }+2v+2 } -\frac { 4 }{ { (v+1) }^{ 2 }+1 } \right] dv=\frac { dx }{ x } \)
Integrating both sides, we get -\(\frac{1}{2}\)log |v²+2v+2|+2tan^-1(v+1) = log |x| + log |C|
or log |v²+2v+2| -4tan^-1(v+1) = -2log |x| -2 log |C|
or log |v²+2v+2| +log |x|² - 4tan^-1(v+1) = -2 log |C|
or log |(v²+2v+2)x²| -4 tan^-1(v+1) = -2 log |C|
Now replacing v by\(\frac{y}{x},\) we get, log |y²+2xy+2x²|-4tan^-1\(\left( \frac { x+y }{ x } \right) =k\), where k = -2 log |C| gives the required solution.