Since, x = a cos(αt + β)
Therefore, dx/dt = a cos(αt + β)
And, d²x/dt² = -a α² cos(αt + β)
= -α² a cos(αt + β)
Or, d²x/dt² = -α² [as a cos(αt + β) = x]
So, d²x/dt² + α²x = 0

The equation of family of parabolas is Ax² + Bx + C = 0 where, A, B, C are arbitrary constant.
By differentiating equation with respect to x till all constants get eliminated,
Hence, d³y/dx³ = 0

dy/dx = (6x + 9y ? 7)/(2x + 3y ? 6)
So, dy/dx = (3(2x + 3y) ? 7)/(2x + 3x ? 6) ???.(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx ? 2)
Therefore, from (1) we get,
1/3(dz/dx ? 2) = (3z ? 7)/(z ? 6)
Or, dz/dx = 2 + (3(3z ? 7))/(z ? 6)
= 11(z ? 3)/(z ? 6)
Or, (z ? 6)/(z ? 3) dz = 11 dx
Or, ∫(z ? 6)/(z ? 3) dz = ∫11 dx
Or, ∫(1 ? 3/(z ? 3)) dz = 11x + c
Or, z ? log |z ? 3| = 11x + c
Or, 2x + 3y ? 11x ? 3log|2x + 3y -3| = c
Or, 3y ? 9x ? 3log|2x + 3y ? 3| = c
Or, 3x ? y + log|2x + 3y ? 3| = -c/3

Consider function y=7x²
Differentiating w.r.t x, we get
\(\frac{dy}{dx}\)=14x
?\(\frac{dy}{dx}\)-14x=0
Hence, function y=7x² is a solution for differential equation \(\frac{dy}{dx}\)-14x=0

dy/dx = (6x + 9y ? 7)/(2x + 3y ? 6)
So, dy/dx = (3(2x + 3y) ? 7)/(2x + 3x ? 6) ???.(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx ? 2)
Therefore, from (1) we get,
1/3(dz/dx ? 2) = (3z ? 7)/(z ? 6)
Or, dz/dx = 2 + (3(3z ? 7))/(z ? 6)
= 11(z ? 3)/(z ? 6)
Or, (z ? 6)/(z ? 3) dz = 11 dx
Or, ∫(z ? 6)/(z ? 3) dz = ∫11 dx
Or, ∫(1 ? 3/(z ? 3)) dz = 11x + c
Or, z ? log |z ? 3| = 11x + c
Or, 2x + 3y ? 11x ? 3log|2x + 3y -3| = c
Or, 3y ? 9x ? 3log|2x + 3y ? 3| = c
Or, 3x ? y + log|2x + 3y ? 3| = -c/3

The given differential equation is polynomial in y’. Therefore, degree of equation will be power raised to highest derivative y’ i.e. 1.

The given equation is a polynomial differential equation. Therefore, degree of equation will power of highest derivative \(\frac{d^2 y}{dx^2}\) i.e. 1.

x(x – 1)dy/dx ? y = x²(x ? 1)²
or, dy/dx ? 1/(x(x ? 1))*y = x(x ? 1) ??(1)
we have, ∫- 1/(x(x ? 1)) dx = – ∫ [1/(x ? 1) ? 1/x] dx
= – [log(x ? 1) ? log x]
= log x ? log(x ? 1)
= log(x/(x ? 1))
Thus, integrating factor = e^{∫- 1/(x(x ? 1)) dx} = x/(x ? 1)
Thus, multiplying both sides of (1) by x/(x ? 1), we get,
x/(x ? 1)*dy/dx ? 1/(x ? 1)² * y = x²
or, d/dx[x/(x ? 1) * y] = x² ??.(2)
integrating both sides of (2) we get,
x/(x – 1) * y = ∫x²dx = x³/3 + c
or xy = (x ? 1)(x³/3 + c)

In D.E y”’-(4y’)³=0, highest order derivative is y”’. Therefore, order of D.E is 3. Since it is a polynomial equation, degree will be power raised to y”’. Therefore, degree is 1.

The original given equation is,
y = ae^3x + be^x ???.(1)
Differentiating above equation, we get
dy/dx = 3ae^3x + be^x ???.(2)
d²y/dx² = 9ae^3x + be^x ???.(3)
Now, to obtain final equation we have to make RHS = 0,
So, to get RHS = 0, we multiply, (1) with 3 and (2) with -4,
Thus , 3y = 3ae^3x + 3be^x ???.(4)
And -4(dy/dx) = -12ae^3x ? 4 be^x ???.(5)
Now, adding above three equations,(i.e. (3) + (4) + (5)) we get,
d²y/dx² ? 4dy/dx + 3y = 0