Ask Me Help Desk - Please help me in solving these questions.Thanks in advance.

Q.1 Solve the given differential equation by Separation of variables.
Sec^2xdy + cscydx = 0

Q.2 Solve the following Differential Equation
(y^2+3xy)dy = (4x^2+xy)dy, y(1) = 1

Q.3 Determine whether the following Differential Equation is exact. If exact, solve.
(1-2x^2-2y)dy/dx = 4x^3 + 4xy

1. I'm not sure about this... It's the first time I see such a differential equation being given like this...

$sec^2 x dy + csc ydx = 0$

$sec^2x dy = -csc y dx$

$\int \frac{1}{-cscy} dy =\int \frac{1}{sec^2x} dx$

$-\int siny\ dy = \int cos^2x\ dx$

You should be able to solve this now.

See here from first time posted:

https://www.askmehelpdesk.com/mathematics/differential-equation-help-needed-521965.html

Right... I didn't notice that it was the one of earlier... :(

Avoid double posting please, mayya and keep to one account.

@Unknown008 yeah that was my mistake in fact. And thanks for the solution.

@Unknown008 that's not me... I have posted this for the first time... that is someone else!I have checked it...

please solve Q.2: (y^2+3xy)dy = (4x^2+xy)dy, y(1) = 1

For Q2, you have a dy on both of them. I believe one should have a dx.

Is this correct:

$(y^{2}+3xy)dx=(4x^{2}+xy)dy$ ?

If this is the correct problem, then it is NOT exact. Use a substitution.

Let $y=ux, \;\ dy=udx+xdu$

After making the subs, simplifying and factoring, we get:

$-ux^{2}(xdu+dx)-4x^{2}(xdu)=0$

Separate variables and integrate:

$\int\frac{u+4}{u}du+\int\frac{1}{x}dx=0$

$4ln(u)+u+ln(x)=C$

Resub $u=\frac{y}{x}$ :

$4ln(\frac{y}{x})+\frac{y}{x}+ln(x)=C$

$4[ln(y)-ln(x)]+\frac{y}{x}+ln(x)=C$

$-3ln(x)+\frac{y}{x}+4ln(y)=C$

Now, use the initial condition, y(1)=1, to find C.

By letting x=1 and y=1, we get C=1.

So, the final solution is $\fbox{-3ln(x)+\frac{y}{x}+4ln(y)=1}$

Here is Q3:

$(1-2x^{2}-2y)\frac{dy}{dx}=4x^{3}+4xy$

Let $M=4x^{3}+4xy$ and $N=2x^{2}+2y-1$

So that $M_{y}=4x=N_{x}$

From $f_{x}=4x^{3}+4xy$ we get, by integrating:

$f=x^{4}+2x^{2}y+h(y), \;\ h'(y)=2y-1$

$h(y)=y^{2}-y$

The solution is then:

$x^{4}+2x^{2}y+y^{2}-y=C$

It helped me a lot. Thank you!
Stay blessed.