Mathematics, 22.09.2019 00:30 ayoismeisalex

Verify that the given differential equation is not exact. (−xy sin(x) + 2y cos(x)) dx + 2x cos(x) dy = 0 if the given de is written in the form m(x, y) dx + n(x, y) dy = 0, one has my = nx = . since my and nx equal, the equation is not exact. multiply the given differential equation by the integrating factor μ(x, y) = xy and verify that the new equation is exact. if the new de is written in the form m(x, y) dx + n(x, y) dy = 0, one has my = nx = . since my and nx equal, the equation is exact. solve.

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Answer from: roseemariehunter12

We are to verify and confirm if the given differential equations are exact or not. Then solve for the exact equation.

The first differential equation says:

$\mathbf{(-xy \ sin x + 2y \ cos x) dx + 2(x \ cos x) dy = 0 }$

Recall that:

A differential equation that takes the form $\mathbf{M(x,y)dt + N(x, y)dy = 0 }$ will be exact if and only if:

$\mathbf{\dfrac{\partial M }{\partial y} = \dfrac{\partial N }{\partial x}}$

From equation (1), we can represent M and N as follows:

$\mathbf{M = (-xy \ sin x + 2y \ cos x)}$ $\mathbf{N = (2x \ cos x)}$

Thus, taking the differential of M and N, we have:

$\mathbf{ \dfrac{\partial M}{\partial y }= M_y = -x sin x + 2cos x}$

$\mathbf{ \dfrac{\partial N}{\partial x }= N_x = 2 cos x + 2x sin x}$

From above, it is clear that:

$\mathbf{\dfrac{\partial M }{\partial y} \neq \dfrac{\partial N }{\partial x}}$

∴

We can conclude that the equation is not exact.

Now, after multiplying the given differential equation in (1) by the integrating factor μ(x, y) = xy, we have:

$\mathbf{ = \mathsf{(-x^2y^2 sin x + 2xy^2cos x ) dx +(2x^2ycos x ) dy = 0 --- (2)}}$

Representing the equation into form M and N, then:

$\mathbf{M = -x^22y^2 sin x +2xy^2 cos x}$

$\mathbf{N = 2x^2y cos x}$

Taking the differential, we have:

$\mathbf{\dfrac{\partial M}{\partial y }= M_y = -2x^2y sin x + 4xy cos x }$

$\mathbf{\dfrac{\partial N}{\partial x} =N_x= 4xycos \ x -2x^2 y sin x}$

Here;

$\mathbf{\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} }$

Therefore, we can conclude that the second equation is exact.

Now, the solution of the second equation is as follows:

$\int_{y } M dx + \int (not \ containing \ 'x') dy = C$

$\rightarrow \int_{y } (-x^2y^2 sin(x) +2xy^2 cos (x) ) dx + \int(0)dy = C$

$\rightarrow-y^2 \int x^2 sin(x) dx +2y ^2 \int x cos (x) dx = C$ ---- (3)

Taking integrations by parts:

$\int u v dx = u \int v dx - \int (\dfrac{du}{dx} \int v dx) dx$

∴

$\int x^2 sin (x) dx = x^2 \int sin(x) dx - \int (\dfrac{d}{dx}(x^2) \int (sin \ (x)) dx) dx$

$\to x^2 (-cos (x)) \ - \int 2x (-cos \ (x)) \ dx$

$\to -x^2 (cos (x)) \ + \int 2x \ cos \ (x) \ dx$ ----- replace this equation into (3)

∴

$\rightarrow-y^2( -x^2 cos (x) \ + \int 2x \ cos \ (x) \ dx) +2y ^2 \int x cos (x) dx = C$

$\mathbf{\rightarrow -x^2 y^2 cos (x) \ -2y ^2 \int x \ cos \ (x) \ dx +2y ^2 \int x cos (x) dx = C}$

$\mathbf{x^2y^2 cos (x) = C\ \text{ where C is constant}}$

Therefore, from the explanation, we've can conclude that the first equation is not exact and the second equation is exact.

Learn more about differential equations here:

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Answer from: chasedehlin3999

The ODE

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

is exact if

$\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$

We have

$M=-xy\sin x+2y\cos x\implies M_y=-x\sin x+2\cos x$

$N=2x\cos x\implies N_x=2\cos x-2x\sin x$

so the ODE is indeed not exact.

Multiplying both sides of the ODE by $\mu(x,y)=xy$ gives

$\mu M=-x^2y^2\sin x+2xy^2\cos x\implies(\mu M)_y=-2x^2y\sin x+4xy\cos x$

$\mu N=2x^2y\cos x\implies(\mu N)_x=4xy\cos x-2x^2y\sin x$

so that $(\mu M)_y=(\mu N)_x$ , and the modified ODE is exact.

We're looking for a solution of the form

$\Psi(x,y)=C$

so that by differentiation, we should have

$\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$

$\implies\begin{cases}\Psi_x=\mu M\\\Psi_y=\mu N\end{cases}$

Integrating both sides of the second equation with respect to gives

$\Psi_y=2x^2y\cos x\implies\Psi=x^2y^2\cos x+f(x)$

Differentiating both sides with respect to gives

$\Psi_x=-x^2y^2\sin x+2xy^2\cos x=2xy^2\cos x-x^2y^2\sin x+\dfrac{\mathrm df}{\mathrm dx}$

$\implies\dfrac{\mathrm df}{\mathrm dx}=0\implies f(x)=c$

for some constant .

So the general solution to this ODE is

$x^2y^2\cos x+c=C$

or simply

$x^2y^2\cos x=C$

Answer from: Quest

i don't have the answer rn but go to math papa. it shows the steps and the answer. hope this : )

step-by-step explanation:

Answer from: Quest

I’m going to infer you said parallel.solve for yy= -1/2x+7/4lines that are parallel have the same slope, different y intercept.so we know m is -1/2then from there solve for y intercept.

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