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$$sec^2(y) \frac{dy}{dx} + 2x \tan(y) = x^3$$ I am not able to get rid of that $dy/dx$ . Please help. PS: I need the answer as $\tan(y) =\frac{x^2 - 1}{2}$

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    Hint: $\tan(y(x))'=y'(x)\sec^2(y(x))$ – 2017-01-24

3 Answers 3

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Put tan y = t

Then we have,

$t' + 2xt = x^3$

It is first order linear equation.

As in the form y' + P(x).y = Q(x)

Solve it to find solution.

u(x) = $e^{\int p(x) dx}$

u(x) = $e^{\int 2x dx}$

u(x) = $e^{x^2}$

Now t = $\frac 1{u(x)} \int u(x). Q(x) dx$

t = $\frac{1}{e^{x^2}} \int e^{x^2}. x^3 dx$

Now solve $\int e^{x^2}. x^3 dx$

Put $x^2 = z$

2x dx = dz

$x dx = \frac12 dz$

Above equation become,

$\frac 12 \int e^z. z dz$

Use integration by parts and replace values.

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    already tried for 2 hours now. I did all other questions of the same kind but this one messes up in later stages thats why I want someone to show me how to do it – 2017-01-24
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    Use the link I send you. Its easy one. – 2017-01-24
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    I know the rules in that file – 2017-01-24
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    But please consider I am preparing for my board exams and do not know higher maths – 2017-01-24
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    kardo paaji please ek question reh gya h please – 2017-01-24
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    Ok yaar me aj krdu pkka. – 2017-01-25
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    Vaise ho gya k nhi? – 2017-01-25
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    nhi hua sir ji . – 2017-01-25
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52442/discussion-between-kanwaljit-singh-and-user5183360). – 2017-01-25
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Hint. Guided by the answer/by inspection, we should make the substitution $z := \tan y$. Then the equation becomes $$ z' + 2xz = x^3$$

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    ok. But what shall I do next to get my answer? – 2017-01-24
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    @user5183360 that would depend on what methods you have covered in class. Strictly speaking since you already know the answer you can just verify that $z = (x^2 - 1)/2$ is a solution and you're done. Did you want more than that? – 2017-01-24
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    I need all the steps to achieve that answer starting from the one in the question. – 2017-01-24
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    @user5183360 Guessing and verifying is a step. If you want one which is more satisfying, what methods did you cover in class? – 2017-01-24
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    Variable seperable. reducable to variable sperable. homogeneous. linear. – 2017-01-24
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    @user5183360 if you've covered integrating factors, see the other answer. – 2017-01-24
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    No I do not know that method. – 2017-01-24
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    Its the simplest one. In this case you just multiply the entire equation by $e^{x^2}$, which turns the LHS into $[ z e^{x^2} ]'$. Another way is to guess that you have a power series solution and then compare coefficients. – 2017-01-24
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    ok I learnt how to get e^x^2 as the integrating factor. The problem is that I can do other questions of this kind but this one messes up at stage. – 2017-01-24
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HINT:

$$\sec^2\left[\text{y}\left(x\right)\right]\cdot\text{y}'\left(x\right)+2\cdot x\cdot\tan\left[\text{y}\left(x\right)\right]=x^3$$

Let $\text{A}\left(x,\text{y}\right)=2\cdot x\cdot\tan\left(\text{y}\right)-x^3$ and $\text{B}\left(x,\text{y}\right)=\sec^2\left(\text{y}\right)$, this is not an exact equation, because:

$$\frac{\partial\space\text{A}\left(x,\text{y}\right)}{\partial\space\text{y}}=2\cdot x\cdot\sec^2\left(\text{y}\right)\ne0=\frac{\partial\space\text{B}\left(x,\text{y}\right)}{\partial x}$$

Find an integration factor, $\rho\left(x\right)$ such that:

$$\rho\left(x\right)\cdot\text{A}\left(x,\text{y}\right)+\rho\left(x\right)\cdot\text{B}\left(x,\text{y}\right)\cdot\text{y}'\left(x\right)=0$$

is exact.

This means:

$$\frac{\partial}{\partial\space\text{y}}\left[\rho\left(x\right)\cdot\text{A}\left(x,\text{y}\right)\right]=\frac{\partial}{\partial x}\left[\rho\left(x\right)\cdot\text{B}\left(x,\text{y}\right)\right]\implies2\cdot x\cdot\rho\left(x\right)\cdot\sec^2\left(\text{y}\right)=\sec^2\left(\text{y}\right)\cdot\rho\left(x\right)$$

This gives:

$$\int\frac{\rho'\left(x\right)}{\rho\left(x\right)}\space\text{d}x=\int2\cdot x\space\text{d}x=\ln\left|\rho\left(x\right)\right|=x^2$$