# P3 Maths Integration Question

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#1
I'M DOING A PST PASER AT THE MOMENT AND I'M HAVING A BIT OF TROUBLE. MAYBE U GUYS COULD HELP ME. THE QUESTION IS QUESTION 3 FROM AN OCR P3 PAPER DATED 11 JUNE 2002 IF THAT HELPS

ANYWAY THE QUESTION IS "FIND THE SOLUTION OF THE DIFFERENTIAL EQUATION: dy/dx = (X*4 - 1)/(x*2 y*2)

given that y=2 when x=1
by the way x*4 means x to the power 4, x*2 means x squared and y*2 means y squared.

i know that the last part is used to find the constant at the end, but the trouble is i can't even intergrate the equation in the first place. can anyone help me with finding the answer. please show ur working.
0
17 years ago
#2
You should "separate the variables". That is, write the DE as

(something involving y only) * dy/dx = (something involving x only).

Then integrate both sides wrt x, using the fact that

(int) [(something involving y only) * dy/dx] dx
= (int) (something involving y only) dy.
0
17 years ago
#3
(Original post by DLo)
I'M DOING A PST PASER AT THE MOMENT AND I'M HAVING A BIT OF TROUBLE. MAYBE U GUYS COULD HELP ME. THE QUESTION IS QUESTION 3 FROM AN OCR P3 PAPER DATED 11 JUNE 2002 IF THAT HELPS

ANYWAY THE QUESTION IS "FIND THE SOLUTION OF THE DIFFERENTIAL EQUATION: dy/dx = (X*4 - 1)/(x*2 y*2)

given that y=2 when x=1
by the way x*4 means x to the power 4, x*2 means x squared and y*2 means y squared.

i know that the last part is used to find the constant at the end, but the trouble is i can't even intergrate the equation in the first place. can anyone help me with finding the answer. please show ur working.
assuming that X=x (see quote),
multiply both sides by y^2 (y^2 means y-squared)
then integral y^2 dy = integral (x^4-1)/x^2 dx
0
17 years ago
#4
y^3=x^3-3/x+10
0
#5
cheers for all ur help guys
it all makes sense now
0
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