Відео

When you multiply ln by negative sign only the sign in front of ln changes. The argument of ln stays the same.

Depends on your engineering program. Some engineers need these : Linear algebra, probability and statistics , numerical analysis as math courses

To recap at the decision point of M or N part of the function we pick one of the ( equally ok) and integrate. So, it will be good choice to pick whichever is easy to integrate. İn my case i picked M part. İntegrated and added some arbitrary function g(y). Different books or parts of school might have their fixed function name. Generally f and g are alternative functions to each other. g(y) has input y, no Mather what function name we picked the input of the arbitrary function is y since we need to solve for y as our main goal. Once i did integral added g(y) and then implicitly differentiated with respect to y to get g'(y). İf there is no y variable in M then they are gone since out implicit is with respect to y. Then i had to set it equal to other part of main equation i did not use. İn our case it is N part. Then g'(y) =N equation. What undoes y in g'(y)! İntegration. Now i get back calculus basics and integrate to obtain y= something else. So you obtain explicit solution.

An exact differential equation is an ODE of the form M(x,y)+N(x,y)y′=0 where there exists a continuously differentiable F(x,y) such that ∂F/∂x=M and ∂F/∂y=N. If M and N are also continuously differentiable, this is equivalent to the condition that ∂M/∂y=∂N/∂x. If the ODE is exact, then by the multi-variable chain rule, d(F(x,y))/dx=∂F/∂x+∂F/∂y*y′=M(x,y)+N(x,y)y′, which means that any solution for y as a function of x has a graph inside a level set of F, a solution to the equation F(x,y)=c for some constant c. Do you mean whether we might have a solution ? You can always express in terms of M if needed or required in the problem. I am a bit confused with the question, sorry :(

Thank you sir❤ ​@@mathbyleo

I mean that the m doesn't have an answer but n does😊

In the implicit form of solution you don't need to solve for "y" yet. You leave the equation as it is. Advantage of implicit for is that you can directly plug in given initial condition values.

😊 hi where are you from

@@Tandiajay you obtain zero "0" because negative pi/2 to pi/2 is symmetric and limits of integration divide s sine fifth into two equal zones that one is positive area and other is negative area. You can graphically check and see.

Ah yes. In the end t supposed to be 4x+6y back. Thanks 👍

Hi, Arslan.

Thank you so much