killjoy1990xb9

2021-12-28

Find the value of k for which the following ODE is an exact ODE $({y}^{3}+kx{y}^{4}-2x)dx+(3x{y}^{2}+20{x}^{2}{y}^{3})dy=0$

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$({y}^{3}+kx{y}^{4}-2x)dx+(3x{y}^{2}+20{x}^{2}{y}^{3})dy=0$

Concept used:

Exact differential equations$Mdx+Ndy=0$ satisfy the condition $M}_{y}={N}_{x$ .

Step 2

Calculation:

If the given differential equation is exact.

Then$M=({y}^{3}+kx{y}^{4}-2x),N=3x{y}^{2}+20{x}^{2}{y}^{3}$

Differentiate$M=({y}^{3}+kx{y}^{4}-2x)$ with respect to y

${M}_{y}=\frac{\partial}{\partial y}({y}^{3}+kx{y}^{4}-2x)$

$=\frac{\partial}{\partial y}\left({y}^{3}\right)+\frac{\partial}{\partial y}\left(kx{y}^{4}\right)-\frac{\partial}{\partial y}\left(2x\right)$

$=3{y}^{2}+4xk{y}^{3}-0$

$=4k{y}^{3}x+3{y}^{2}$

Differentiate$N=3x{y}^{2}+20{x}^{2}{y}^{3}$ with respect to x.

${N}_{x}=\frac{\partial}{\partial x}(3x{y}^{2}+20{x}^{2}{y}^{3})$

$=\frac{\partial}{\partial x}\left(3x{y}^{2}\right)+\frac{\partial}{\partial x}\left(20{x}^{2}{y}^{3}\right)$

$=3{y}^{2}+40{y}^{3}x$

Step 3

If the differential equation is exact then$M}_{y}={N}_{x$

$4k{y}^{3}x+3{y}^{2}=3{y}^{2}+40{y}^{3}x$

$4k{y}^{3}x=40{y}^{3}x$

$k=10$

The value of$k=10$ .

(a) To find: The value of k for which the given ODE is an exact ODE.

Given information:

The ordinary differential equation is

Concept used:

Exact differential equations

Step 2

Calculation:

If the given differential equation is exact.

Then

Differentiate

Differentiate

Step 3

If the differential equation is exact then

The value of

Neil Dismukes

Beginner2021-12-30Added 37 answers

Comparing this with:

$P(x,y)dx+Q(x,y)dy=0$

We get;

$P(x,y)={y}^{3}+kx{y}^{4}-2x;\text{}Q(x,y)=3x{y}^{2}+20{x}^{2}{y}^{3}$

Now we compute the partial derivatives:

$P}_{y}=3{y}^{2}+4kx{y}^{3$

$Q}_{x}=3{y}^{2}+40x{y}^{3$

Since it is given that the given differential equation is exact:

$P}_{y}={Q}_{x$

$3{y}^{2}+4kx{y}^{3}=3{y}^{2}+40x{y}^{3}$

$4kx{y}^{3}=40x{y}^{3}$

$k=10$

We get;

Now we compute the partial derivatives:

Since it is given that the given differential equation is exact:

Vasquez

Expert2022-01-09Added 669 answers

Therefore,

Solve for k

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