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1. There is a differential equation :dy/dx = (x^2+ 3y^2)/2xy
2. How to solve the differential equation above and find the solution of the differential equation ?
There is an equation :
Z = 2e^3y cos⁡2x and point P( π/3, 0, -1 )
Determine the tangent plane equation toward the surface of the point above !
3. In a mathematics test, each students can choose 8 questions from 10 equations. If number 4 and 9 should be solve (do), how many choice questions that students can do?
4. Determine mac laurin series for function f(x) = cos (x2) !
5. Obserb whether these line and plane parallel or not ?
h≡(■(x,y,z))=(■(0,4,0))+t(■(5,-2,-2))α≡(■(x,y,z))=(■(2,1,0))+a(■(-1,2,0))+b(■(2,0,-1))

Solving :
1. dy/dx = (x^2+ 3y^2)/2xy the form of homogeneous equation suggest that it may be simplified by introducing a new variable, which will we denote by v, to represent the ratio of y to x. Thus : y = xv ↔ v = y/x
And y = xv becomes dy/dx = x dv/dx+ v
Writing dy/dx = (x^2+ 3y^2)/2xy as dy/dx = (x^2 (1+ 3 y^2/x^2 ))/(x^2 (2 y/x) )
dy/dx = ((1+ 3 y^2/x^2 ))/((2 y/x) )
It shows that this equation is homogenous. Then given equation is neither linear, nor separable, nor exact. So we can substitute dy/dx = x dv/dx+ v into dy/dx = ((1+ 3 y^2/x^2 ))/((2 y/x) ) we obtain :
x dv/dx+ v = ((1+ 3v^2 ))/((2v) )
x dv/dx = ((1+ 3v^2 ))/((2v) )-v
x dv/dx = (1+3v^2-2v^2)/2v
x dv/dx = (1+v^2)/2v
if v ≠ 0, then equation above can be written as 2v/(1+v^2 ) dv= 1/x dx
integrating both sides, assumption : u=1+v^2 and du=2v dv
so , ∫▒〖u^(-1) du= ∫▒1/x〗 dx
〖 ln〗⁡|u|+ln|c_1 |=ln|x|+ln|c_2 |
〖 ln〗⁡|1+v^2 |+ln|c_1 |=ln|x|+ln|c_2 | ↔ 〖 ln〗⁡|1+v^2 |= ln|x|+ln|c|
Where c is an orbitrary constant. Hence, combining the logarithms and taking the exponential of both sides, we obtain :
e^〖 ln〗⁡|1+v^2 | =e^(ln|x|+ln|c| )
1+v^2=cx
v^2=cx-1
Finally, substituting for v in terms of y gives the solution of dy/dx = (x^2+ 3y^2)/2xy in the form :
y^2/x^2 =cx-1
y^2=cx^3-x^2
So, y=√(cx^3-x^2 )

2. Z = 2e^3y cos⁡2x and point P( π/3, 0, -1 )
Assumption : f(x,y)=2e^3y cos⁡2x then,
∇f(x,y)=-4e^3y cos⁡〖2xi+6e^3y cos⁡2xj 〗
Substitution (π/3,0) to equation above
∇f(x,y)=-4e^3.0 cos⁡〖2(π/3)i+6e^3.0 cos⁡2(π/3)j 〗
=-2√3 i-3j
In accordance with theorem the tangent plane equation at ( π/3, 0, -1 ) is
z+1=-2√3 (x-π/3)-3(y-0)
z+1=-2√3 x+2π/√3-3y
2√3 x+3y+z=2π/√3-1
3. Each students can choose 8 questions from 10 questions. Number 4 and 9 should be solve, so each students can choose 6 questions. With used combination formula, we obtain :
C_(8-2)^(10-2)=C_6^8=8!/(8-6)!6!= 8.7.6!/2.6!= 28
So, many ways to choose the questions are 28 ways.

4. f(x)=cos⁡〖(x^2 〗)→f(0)=1
f^' (x)=-2 sin⁡(x^2 )→f^' (0)=0
f^'' (x)=-4 cos⁡〖(x^2 〗) →f^'' (0)=-4
f^''' (x)=8sin⁡(x^2 ) →f^''' (0)=0
f^IV (x)=16cos⁡〖(x^2 〗) →f^IV (0)=16
f^v (x)=-32 sin⁡(x^2 ) →f^v (0)=0
f^VI (x)=-64 cos⁡〖(x^2 〗) → f^VI (0)=-64,etc
So, mac laurin series for f(x)=cos⁡〖(x^2 〗) is
cos⁡〖(x^2 〗)=f(0)+f^' (0)x+(f"(0)x^2)/2!+(f"'(0)x^3)/3!+ …..
cos⁡〖(x^2 〗)=1+0.x+(-4x^2)/2!+(0.x^3)/3!+(16.x^4)/4!+(0.x^5)/5!+(-64.x^6)/6!+⋯
cos⁡〖(x^2 〗)=1-2x^2+(2.x^4)/3-(4.x^6)/15+…..

5. h≡(■(x,y,z))=(■(0,4,0))+t(■(5,-2,-2))
α≡(■(x,y,z))=(■(2,1,0))+a(■(-1,2,0))+b(■(2,0,-1))
Gradient of h and α form become matriks
Assumption=A=[■(5,-2,-2),(-1,2,0),(2,0,-1)]
Then determine determinant A
Det(A) = (■((5,-2,-2),(-1,2,0),(2,0,-1)│■((5,-2,-2),(-1,2,0))=0
Because determinant A equals to zero, so h lines parallel with α plane.