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I have been solving this ODE

$ydx+(y+\tan(x+y))dy=0$

My approach was like this;

As this equation is not exact;

$\frac{\partial}{\partial y}y = 1$

$\frac{\partial}{\partial x}(y+\tan(x+y)) = \sec^{2}(x+y)$

We have to make this ODE an exact equation.

Thus, I thought that as there is $\tan(x+y)$, we cannot integrate both side of this equation(for there might be $\log\mid\cos(x+y)\mid$ form), I thought I might have to use the property of the trigometric function

$\frac{d}{dx}\sin x = \cos x$

$\frac{d}{dx}\cos x = -\sin x$

So I multiplied $\cos (x+y)$ at both side of this ODE and got

$y\cos(x+y)dx + (y\cos(x+y) + \sin(x+y))dy = 0$

Which is exact;

$\frac{\partial}{\partial y}y\cos(x+y) = \cos(x+y) - y\sin(x+y)$

$\frac{\partial}{\partial x}(y\cos(x+y) + \sin(x+y)) = - y\sin(x+y)+\cos(x+y)$

And with some more steps, I finally got answer

$y\sin(x+y) = c$

Where $c$ is constant.

But here's my question.

I tried to find out the integrating factor $F(x,y)$ which makes

$Fydx+F(y+\tan(x+y))dy=0$

an exact equation. This $F(x,y)$ should satisfy

$y\frac{\partial F}{\partial y} + F = \frac{\partial F}{\partial x}(y+\tan(x+y))+F\sec^{2}(x+y)$

But I cannot move forward form this step. I can assure that this $F(x, y)$ is exactly $\cos(x+y)$, but I don't know how to construct it or prove it.

Is there any help that I can find out integrating factor of this ODE?

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Assume that you have an ode of the form $P(x,y)\text{d}x+Q(x,y)\text{d}y=0$ which is not exact.

We want to find an integrating factor of the form $\mu(x+y)$, i.e the equation $$\mu(x+y)P(x,y)\text{d}x+\mu(x+y)Q(x,y)\text{d}y=0$$ is exact. Define $\mu(x+y)P(x,y)=M(x,y),\mu(x+y)Q(x,y)=N(x,y)$.

The equation is exact, hence $M_y=N_x$, thus we have $$\mu'(x+y)P(x,y)+\mu(x+y)P_y=\mu'(x+y)Q(x,y)+\mu(x+y)Q_x$$Eventually $$\frac{\mu'(x+y)}{\mu(x+y)}=\frac{Q_x-P_y}{P(x,y)-Q(x,y)}$$

In your case, $Q_x=\sec^2(x+y),P_y=1$ and we get $$\frac{\mu'(x+y)}{\mu(x+y)}=\frac{\sec^2(x+y)-1}{-\tan(x+y)}=-\tan(x+y)\Longrightarrow \ln\left[\mu(x+y)\right]=\ln\left[\cos(x+y)\right]$$

Thus, the integrating factor is $\mu(x+y)=\cos(x+y)$.

The hint of finding an integrating factor that depends on $x+y$ is the function $Q(x,y)$.

You might want to try this and find integrating factor of the forms $\displaystyle \mu(xy),\mu\left(\frac{x}{y}\right)$, etc.

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