Now solve it
Mayank and Akul have been busy with their JEE preparation so Mayank thought of adding a small question without much story line
Solve
d x d y = x − 2 x 2 y 3 3 x y 4 + 2 y
where x = 0 , you're given that y = 1 when x = − 1 . Find the value of x when y = − 0 . 5 .
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3 solutions
The solution won't be complete if you won't tell what inspired you towards the very first step of your solution
Very elegant approach! . but how did you thought of it @Jon Haussmann
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First, I wrote the equation in the form ( 3 x y 4 + 2 y ) + d x d y ( 2 x 2 y 3 − x ) = 0 . I then multiplied both sides by an appropriate factor to get an exact differential equation; in this case, the factor is y 2 x .
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@Jon Haussmann Sir, I assumed that the appropriate integrating factor is of the form x m y n but after this, I found after simplification that 3 n − 2 m + 8 = 0 but could not proceed further...........any help???
For reaching the above equation I mainly resorted to the checking the condition for the expression being an exact differential....
After multiplying both sides by x m y n , we get ( 3 x m + 1 y n + 4 + 2 x m y n + 1 ) + d x d y ( 2 x m + 2 y n + 3 − x m + 1 y n ) = 0 .
Note that d y d ( 3 x m + 1 y n + 4 + 2 x m y n + 1 ) = 3 ( n + 4 ) x m + 1 y n + 3 + 2 ( n + 1 ) x m y n and d x d ( 2 x m + 2 y n + 3 − x m + 1 y n ) = 2 ( m + 2 ) x m + 1 y n + 3 − ( m + 1 ) x m y n . Thus, we can make the differential equation exact by finding m and n such that 3 ( n + 4 ) = 2 ( m + 2 ) and 2 ( n + 1 ) = − ( m + 1 ) . This gives us m = 1 and n = − 2 .
Relevant wiki: Variable changes in linear differential equations of first order: y'=f(t,y)
Set xy^3 = v . i thought of this substitution as on taking y and x common from numerator and denominator respectively i saw the terms of xy^3 coming .
On a bit of simplification you will obtain
7dy/y = 2dv/v +dv/(v+1)
Integrating both sides and finding value of arbitary constant of integration using initial conditions we will get differential equation as
x^2(xy^2 + 1/y) = 0
Bhaiya i wanted to know what rank did you got in AITS? In mains and advance
Well...we could solve it by three subsequent substitutions.... Put (y^3= t) Follow it by replacing t by 1/t Third one..the standard method of solving homogeneous equations
We can re-write the differential equation as d x d ( x 3 y 2 + y x 2 ) = 0 . Thus, x 3 y 2 + y x 2 = c for some constant c . Setting x = − 1 and y = 1 , we get c = 0 . For y = − 1 / 2 , we get 4 x 3 − 2 x 2 = 4 x 2 ( x − 8 ) = 0 . Since x = 0 , x = 8 .