killjoy1990xb9

2021-12-28

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

Thomas White

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 $\left({y}^{3}+kx{y}^{4}-2x\right)dx+\left(3x{y}^{2}+20{x}^{2}{y}^{3}\right)dy=0$
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 $M=\left({y}^{3}+kx{y}^{4}-2x\right),N=3x{y}^{2}+20{x}^{2}{y}^{3}$
Differentiate $M=\left({y}^{3}+kx{y}^{4}-2x\right)$ with respect to y
${M}_{y}=\frac{\partial }{\partial y}\left({y}^{3}+kx{y}^{4}-2x\right)$
$=\frac{\partial }{\partial y}\left({y}^{3}\right)+\frac{\partial }{\partial y}\left(kx{y}^{4}\right)-\frac{\partial }{\partial y}\left(2x\right)$
$=3{y}^{2}+4xk{y}^{3}-0$
$=4k{y}^{3}x+3{y}^{2}$
Differentiate $N=3x{y}^{2}+20{x}^{2}{y}^{3}$ with respect to x.
${N}_{x}=\frac{\partial }{\partial x}\left(3x{y}^{2}+20{x}^{2}{y}^{3}\right)$
$=\frac{\partial }{\partial x}\left(3x{y}^{2}\right)+\frac{\partial }{\partial x}\left(20{x}^{2}{y}^{3}\right)$
$=3{y}^{2}+40{y}^{3}x$
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 $k=10$.

Neil Dismukes

Comparing this with:
$P\left(x,y\right)dx+Q\left(x,y\right)dy=0$
We get;

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:
${P}_{y}={Q}_{x}$
$3{y}^{2}+4kx{y}^{3}=3{y}^{2}+40x{y}^{3}$
$4kx{y}^{3}=40x{y}^{3}$
$k=10$

Vasquez

Therefore, $3{y}^{2}+4kx{y}^{3}=3{y}^{2}+40x{y}^{3}$
Solve for k $4kx{y}^{3}=40x{y}^{3}$
$k=\frac{40x{y}^{3}}{4x{y}^{3}}$
$k=10$

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