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DE: $$y'\cos x+2y\sin x=0$$ IC:$$y(\pi/2)=1/2$$

Here is my work so far:

Step 1 (divide the equation by $cosx$ and arrange the terms): $$\frac{dy}{dx}=-2y\tan x$$

Step 2 (separate the variables and integrate): $$\int\frac{dy}{-2y}=\int\frac{\sin x}{\cos x}$$

Step 3: $$-\frac{1}{2}\ln y=-\ln|\cos x|+C$$

Step 4 (Plugging in the Initial Condition and find C): $$-\frac{1}{2}\ln(\frac{1}{2})=-\ln|\cos(\frac{\pi}{2})|+C$$

The problem now is $\ln0$ is on the right side. Did I solve the DE wrong?

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    i have the solution $$y(x)=C\cos(x)^2$$ i think it is the same – 2017-01-30
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    What is C in your case? – 2017-01-30
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    Your **IC** doesn't satisfy your **DE** – 2017-01-30
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    but since $$\cos(\frac{\pi}{2})=0$$ your Logarithmus function is not defined – 2017-01-30
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    The IC is for the solution not the DE. – 2017-01-30
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    Not different. It must to satisfy. – 2017-01-30
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    The IC don't satisfy the DE, ergo, there can't be any derivable function that is solution of the DE and also that satisfices the IC – 2017-01-30
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    To repeat what others already mentioned, perhaps more explicitly, if indeed $$y'(x)\cos(x)+2y(x)\sin(x)=0$$ for every $x$ in an unspecified interval containing $x_0=\frac\pi2$ and if $$y(x_0)=\tfrac12$$ then $$1=y'(x_0)\cdot0+2\cdot\tfrac12\cdot1=0$$ which is absurd. Hence this differential equation has no solution such that $y(x_0)=\frac12$ (and actually, no solution such that $y(x_0)\ne0$). – 2017-02-01

1 Answers 1

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Well, we have:

$$\text{y}'\left(x\right)\cdot\cos\left(x\right)+2\cdot\text{y}\left(x\right)\cdot\sin\left(x\right)=0\space\Longleftrightarrow\space\int\frac{\text{y}'\left(x\right)}{\text{y}\left(x\right)}\space\text{d}x=\int\frac{-2\cdot\sin\left(x\right)}{\cos\left(x\right)}\space\text{d}x\tag1$$

Now, use:

  • $$\int\frac{\text{y}'\left(x\right)}{\text{y}\left(x\right)}\space\text{d}x=\ln\left|\text{y}\left(x\right)\right|+\text{C}_1\tag2$$
  • $$\int\frac{-2\cdot\sin\left(x\right)}{\cos\left(x\right)}\space\text{d}x= -2\int\tan\left(x\right)\space\text{d}x=2\ln\left|\cos\left(x\right)\right|+\text{C}_2\tag3$$

So, we get that:

$$\ln\left|\text{y}\left(x\right)\right|=2\cdot\ln\left|\cos\left(x\right)\right|+\text{C}\tag4$$

When we want to solve $\text{C}$, use the initial condition $\text{y}\left(\frac{\pi}{2}\right)=\frac{1}{2}$, but we get:

$$\cos\left(\frac{\pi}{2}\right)=0\tag5$$

And, when we know that:

$$\lim_{x\to\frac{\pi}{2}}\ln\left|\cos\left(x\right)\right|=\lim_{\text{n}\to0}\ln\text{n}\space\space\space\to\space\space\space-\infty\tag6$$

And so, we get:

$$\ln\left|\frac{1}{2}\right|=-\ln(2)\ne\lim_{x\to\frac{\pi}{2}}\ln\left|\cos\left(x\right)\right|=\lim_{\text{n}\to0}\ln\text{n}\space\space\space\to\space\space\space-\infty\tag7$$

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    So the conclusion is the two sides don't equal? Why? – 2017-01-30
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    That is strange. Maybe the wrong initial condition? – 2017-01-30
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    @user132522 because $-\ln(2)\approx-0.6931471806$ and that does not tends toward $-\infty$. So when you want to solve $\text{C}$ it will not be a 'number'! So, we can conclude that when we want to solve: $$\ln\left|\frac{1}{2}\right|=-\ln(2)=2\cdot\ln\left|\cos\left(\frac{\pi}{2}\right)\right|+\text{C}$$ there are no solutions for this equation!!! Because we will something like: $$-\ln(2)=2\cdot\left(-\infty\right)+\text{C}$$ And that is really strange (also keep in mind that $-\infty$ or $\infty$ is not defined as a number, so my notation is strange and only to let you see what happend) – 2017-01-31
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    @user132522 Yes, that is the only option left!! – 2017-01-31
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    This repeats the content of the OP and avoids the question asked. – 2017-02-01
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    ?? Well, yes, indeed a factual argument. You should know by now that offtopic and/or actually wrong answers are routinely accepted on the site (your own collection of answers being as good as any database of the phenomenon). – 2017-02-04