The given is a polynomial differential equation in \(\frac{d^3 y}{dx^3}\). Therefore, its degree will be power raised to highest order derivative \(\frac{d^3 y}{dx^3}\) which is 1.

Let x cm be distance of moving particle from origin and v cm/sec be its velocity at end of t seconds. Then acceleration of particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t² ? 5t
Or, dv = 3t² dt ? 5t dt
Or, ∫dv = 3∫t² dt ? 5∫t dt
Or, v = t³ ? (5/2)t² + c ???.(1)
Given, v = 5, when t = 0; hence putting se values in equation (1) we get, c = 5
Thus v = t³ ? (5/2)t² + 5
Or, dx/dt = t³ ? (5/2)t² + 5 ???..(2)
Or, dx = t³ dt ? (5/2)t² dt + 5 dt
Integrating this we get,
x = (1/4)t⁴ ? (5/2)t³/3 + 5t + k ??….(3)
By problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t⁴ ? (5/6)t³ + 5t ???.(4)
Thus, velocity of particle at end of 4 seconds,
= [x]_{t = 4} = (1/4)4⁴ ? (5/6)4³ + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm

The highest order derivative in equation is y”’. Hence, order is 3. The equation is polynomial in y”’. Therefore, degree of D.E will be power of derivative y”’ i.e. 2.

The number of arbitrary constants in a general solution of a n^th order differential equation is n.
Therefore, number of arbitrary constants in general solution of a second order D.E is 2.

dy/dx ? a/x * y = (x + 1)/x ??.(1)
Multiplying both sides of equation (1) by
e^∫-a/xdx
= e^{-a log x}
= e^{log x-a}
= x^-a
We get, x^-ady/dx – x^-a (a/x)y = x^-a (x + 1)/x
Or, d/dx(y . x^-a) = x^-a + x^{-a ? 1} ??.(2)
Integrating both sides of (2) we get,
y. x^-a = x^{-a + 1}/(-a + 1) + x^{-a ? 1 + 1}/(-a -1 + 1) + c
= x^-a.x/(1 ? a) + x^-a/-a + c
Or, y = x/(1 ? a) ? 1/a + cx^a

In differential equation 3y”-y’-e^y=0, highest order derivative is y”. Therefore, order is 2. The D.E is not polynomial, so degree of differential equation is not defined.

Given, d²y/dx² = e^2x(12 cos3x ? 5 sin3x) ???(1)
Integrating (1) we get,
dy/dx = 12∫ e^2xcos3x dx – 5∫ e^2x sin3x dx
= 12 * (e^2x/2² + 3²)[2cos3x + 3sin3x] ? 5 * (e^2x/2² + 3²)[2sin3x ? 3cos3x] + A (A is integrationconstant)
So dy/dx = e^2x/13 [24 cos3x + 36 sin3x ? 10 sin3x + 15 cos3x] + A
= e^2x/13(39 cos3x + 26 sin3x) + A
=> dy/dx = e^2x(3 cos3x + 2 sin3x) + A ???.(2)
Again integrating (2) we get,
y = 3*∫ e^2x cos3xdx + 2∫ e^2x sin3xdx + A ∫dx
y = 3*(e^2x/2² + 3²)[2cos3x + 3sin3x] + 2*(e^2x/2² + 3²)[2sin3x ? 3cos3x] + Ax + B (B is integration constant)
y = e^2x/13(6 cos3x + 9 sin3x + 4 sin3x ? 6 cos3x) + Ax + B
or, y = e^2x sin3x + Ax + B

The given differential equation is not polynomial. Hence, degree of differential equation will not be defined.

The equation of tangent to curve y = f(x), at point (x, y), is
Y ? y = dy/dx * (X ? x) ?..(1)
Where it meets x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets y axis, X = 0 and Y = (y – x/(dy/dx))
Also, area of triangle formed by (1) with coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))² ? 4dy/dx = 0
Or, x²(dy/dx)² ? 2(xy ? 2)dy/dx + y² = 0
Solving for dy/dx we get,
dy/dx = [(xy ? 2) ± √(1 ? xy)]/ x²
Let, 1 ? xy = t²
=> x(dy/dx) + y = -2t(dt/dx)
=> x²(dy/dx) = t² ? 1 ? 2tx(dt/dx), so that (3) gives
t(x(dt/dx) ? (t ± 1)) = 0
Hence, eir t = 0
=>xy = 1 which is satisfied by (1, 1)
Or, x dt/dx = t ± 1
=> dx/x = dt/t ± 1
=> t ± 1 = cx
For x = 1, y = 1 and t = 0
=> c = ± 1, so solution is
t = ± (x ? 1) => t² = (x ? 1)²
Or, 1 ? xy = x² ? 2x + 1
Or, x + y = 2
Thus, two curves that satisfies are xy = 1 and x + y = 2

Given that, \(\frac{dy}{dx}=\frac{y-3}{x-3}\)
Separating variables, we get
\(\frac{dy}{y-3}=\frac{dx}{x-3}\)
log?(y-3)=log?(x-3)+log?C₁
log?(y-3)-log?(x-3)=log?C₁
\(log?(\frac{y-3}{x-3})\)=log?C₁
\(\frac{1}{C_1} \frac{y-3}{x-3}=0\)
y-3=0 is general solution for given differential equation.