Find the value of k for which the following ODE

killjoy1990xb9

killjoy1990xb9

Answered question

2021-12-28

Find the value of k for which the following ODE is an exact ODE (y3+kxy42x)dx+(3xy2+20x2y3)dy=0

Answer & Explanation

Thomas White

Thomas White

Beginner2021-12-29Added 40 answers

Step 1
(a) To find: The value of k for which the given ODE is an exact ODE.
Given information:
The ordinary differential equation is (y3+kxy42x)dx+(3xy2+20x2y3)dy=0
Concept used:
Exact differential equations Mdx+Ndy=0 satisfy the condition My=Nx.
Step 2
Calculation:
If the given differential equation is exact.
Then M=(y3+kxy42x),N=3xy2+20x2y3
Differentiate M=(y3+kxy42x) with respect to y
My=y(y3+kxy42x)
=y(y3)+y(kxy4)y(2x)
=3y2+4xky30
=4ky3x+3y2
Differentiate N=3xy2+20x2y3 with respect to x.
Nx=x(3xy2+20x2y3)
=x(3xy2)+x(20x2y3)
=3y2+40y3x
Step 3
If the differential equation is exact then My=Nx
4ky3x+3y2=3y2+40y3x
4ky3x=40y3x
k=10
The value of k=10.
Neil Dismukes

Neil Dismukes

Beginner2021-12-30Added 37 answers

Comparing this with:
P(x,y)dx+Q(x,y)dy=0
We get;
P(x,y)=y3+kxy42x; Q(x,y)=3xy2+20x2y3
Now we compute the partial derivatives:
Py=3y2+4kxy3
Qx=3y2+40xy3
Since it is given that the given differential equation is exact:
Py=Qx
3y2+4kxy3=3y2+40xy3
4kxy3=40xy3
k=10
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

M(x,y)dx+N(x,y)dy=0(y3+kxy42x)dx+(3xy2+20x2y3)dy=0My=y[y3+kxy42x]=3y2+4kxy3Nx=x[3xy2+20x2y3]=3y2+40xy3My=Nt
Therefore, 3y2+4kxy3=3y2+40xy3
Solve for k 4kxy3=40xy3
k=40xy34xy3
k=10

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