Test whether the differential equations are exact and solve those

Francisca Rodden

Francisca Rodden

Answered question

2021-12-29

Test whether the differential equations are exact and solve those that are 3x2y2dx+(2xy3+4y3)dy=0

Answer & Explanation

amarantha41

amarantha41

Beginner2021-12-30Added 38 answers

Given differential equation
3x2y2dx+(2xy3+4y3)dy=0 (1)
Comparing (1) with Mdx+Ndy=0 we get
M=3x2y2,N=2x3x+4y3 Now My=y(3x2y2)=3x2y(y2)=3x2×2y=6x2y
Nx=x(2x3y+4y3)=2yx(x3)+x(4y3)
=2y×3x2+0
=6x2y
Since My=Nx=6x2y
So the differential equation (1) is exact
Since the given differential equation exact so the solution
Mdx(y=constant)+N(the term not containedx)dy=c
3x2y2dx+4y3dy=C
y=constant
3y2x2dx+4y44=c
3y2×x33+4y44=c
y2x3+y4=c
Hence the solution is y2x3+y4=c
Annie Gonzalez

Annie Gonzalez

Beginner2021-12-31Added 41 answers

Simplifying
3x2y2dx+(2x3y+4y3)dy=0
Multiply x2y2dx
3dx3y2+(2x3y+4y3)dy=0
Reorder the terms for easier multiplication:
3dx3y2+dy(2x3y+4y3)=0
3dx3y2+(2x3ydy+4y3dy)=0
3dx3y2+(2dx3y2+4dy4)=0
Combine like terms: 3dx3y2+2dx3y2=5dx3y2
5dx3y2+4dy4=0
Solving
5dx3y2+4dy4=0
Solving for variable 'd'.
Move all terms containing d to the left, all other terms to the right.
Factor out the Greatest Common Factor (GCF), 'dy2'.
dy2(5x3+4y2)=0
karton

karton

Expert2022-01-09Added 613 answers

3x2y2dx+(2xy3+4y3)dy=0M=3x2y2,N=2x3x+4y3Now My=y(3x2y2)=3x2y(y2)=3x2×2y=6x2yNx=x(2x3y+4y3)=2yx(x3)+x(4y3)=2y×3x2+0=6x2yMy=Nx=6x2yMdx(y=constant)+N(the term not containedx)dy=c3x2y2dx+4y3dy=Cy=constant3y2x2dx+4y44=c3y2×x33+4y44=cy2x3+y4=cy2x3+y4=c

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