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The problem is:

Solve the ordinary differential equation:

$$\tan(y) \;dy-\left(\frac{e^{(2x)}\cdot\cos(y+3)\cdot \sqrt x}{2}\right) \cos(y)\; dx=0$$

I am trying to find the solution of this by using an integrating factor or linear equation, but not get anything.

  • 0
    I edited the formatting of the equation by adding parenthesis. Check if that's what you want to express. – 2017-02-07
  • 1
    @Psuresh: **Hint:** It is a Separable Equation - a bit of an ugly result. – 2017-02-07

1 Answers 1

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Well, we know that:

$$\tan\left(\text{y}\right)\space\text{d}\text{y}-\left(\frac{\cos\left(3+\text{y}\right)\cdot e^{2x}\cdot\sqrt{x}}{2}\right)\cdot\cos\left(\text{y}\right)\space\text{d}x=0\tag1$$

This gives that:

$$2\cdot\frac{\text{y}'\left(x\right)}{\cos\left(\text{y}\left(x\right)\right)\cdot\cos\left(3+\text{y}\left(x\right)\right)\cdot\cot\left(\text{y}\left(x\right)\right)}=e^{2x}\cdot\sqrt{x}\tag2$$

Take the integral of both sides with respect to $x$:

  1. For the LHS, substitute $\text{u}=\text{y}\left(x\right)$: $$\mathcal{I}=\int2\cdot\frac{\text{y}'\left(x\right)}{\cos\left(\text{y}\left(x\right)\right)\cdot\cos\left(3+\text{y}\left(x\right)\right)\cdot\cot\left(\text{y}\left(x\right)\right)}\space\text{d}x=$$ $$2\int\frac{1}{\cos\left(\text{u}\right)\cdot\cos\left(3+\text{u}\right)\cdot\cot\left(\text{u}\right)}\space\text{d}\text{u}\tag3$$

Now, substitute $\text{v}=\tan\left(\text{u}\right)$:

$$\mathcal{I}=2\int\frac{\text{v}}{\text{v}\sin\left(3\right)-\cos\left(3\right)}\space\text{d}\text{v}=2\cot\left(3\right)\int\frac{1}{\text{v}\sin\left(3\right)-\cos\left(3\right)}\space\text{d}\text{v}+2\csc\left(3\right)\int1\space\text{d}\text{v}=$$ $$2\cot\left(3\right)\int\frac{1}{\text{v}\sin\left(3\right)-\cos\left(3\right)}\space\text{d}\text{v}+2\csc\left(3\right)\ln\left|\text{v}\right|\tag4$$

Now, substitute $\text{w}=\text{v}\sin\left(3\right)-\cos\left(3\right)$:

$$2\cot\left(3\right)\int\frac{1}{\text{v}\sin\left(3\right)-\cos\left(3\right)}\space\text{d}\text{v}=2\cot\left(3\right)\csc\left(3\right)\int\frac{1}{\text{w}}\space\text{d}\text{w}=2\cot\left(3\right)\csc\left(3\right)\ln\left|\text{w}\right|+\text{C}_1\tag5$$

  1. For the RHS: $$\int e^{2x}\cdot\sqrt{x}\space\text{d}x=\frac{4e^{2x}-\sqrt{2\pi}\cdot\text{erfi}\left(\sqrt{2}\cdot\sqrt{x}\right)}{8}+\text{C}_2\tag6$$

So, we get that:

$$2\csc\left(3\right)\left(\cot\left(3\right)\ln\left|\tan\left(\text{y}\left(x\right)\right)\sin\left(3\right)-\cos\left(3\right)\right|+\ln\left|\tan\left(\text{y}\left(x\right)\right)\right|\right)=\frac{4e^{2x}-\sqrt{2\pi}\cdot\text{erfi}\left(\sqrt{2}\cdot\sqrt{x}\right)}{8}+\text{C}$$