Solve the given first order differential equation

[chwala]

Homework Statement

This is a text question-

Relevant Equations

separation of variables
first order differential equation- integrating factor...

My thinking is two-fold, firstly, i noted that we can use separation of variables; i.e

##\dfrac{dy}{y}= \sec^2 x dx##

on integrating both sides we have;

##\ln y = \tan x + k##

##y=e^{\tan x+k} ##

now i got stuck here as we cannot apply the initial condition ##y(\dfrac {π}{4})=-1##

Secondly on using;

##\dfrac{dy}{dx}+ P(x)y=q(x)##

i have

##\dfrac{dy}{dx}-\sec^2 x=0##

i.f= ##e^{-\int sec^2x dx} =e^{-\tan x}##

therefore,

##(e^{-\tanx }⋅y)' =0## on integration, we shall have;

##(e^{-\tan x} ⋅y) =k## now using the initial condition, ##y(\dfrac {π}{4})=-1##

we have, ##k=-\dfrac{1}{e}##

thus,
##y=e^{\tan x} ⋅k=\left[ e^{\tan x} ⋅-\dfrac{1}{e}\right]=-e^{\tan x-1}##

i do not have solution to the problem...your insight is welcome...

 

[phyzguy]

I don't see what your problem is. You found that $$y=-e^{\tan(x) -1}$$

Isn't this answer (B)?

 

[Mark44]

on integrating both sides we have;
##\ln y = \tan x + k##

##y=e^{\tan x+k} ##

now i got stuck here as we cannot apply the initial condition ##y(\dfrac {π}{4})=-1##

Your second equation above is equivalent to ##y = e^{\tan(x)}\cdot e^k## or ##y = Ce^{\tan(x)}##, where ##C= e^k##. Simply substitute ##\pi/4## for x and -1 for y to solve for C.

 

[SammyS]

Homework Statement:: This is a text question-
Relevant Equations:: separation of variables
first order differential equation- integrating factor...

View attachment 319644

My thinking is two-fold, firstly i noted that we can use separation of variables; i.e

##\dfrac{dy}{y}= \sec^2 x dx##

on integrating both sides we have;

##\ln y = \tan x + k##

##y=e^{\tan x+k} ##

now i got stuck here as we cannot apply the initial condition ##y(\dfrac {π}{4})=-1##

Keep in mind:

##\displaystyle \int \dfrac{dy}{y} = \ln |y| + C ##

 

[chwala]

I don't see what your problem is. You found that $$y=-e^{\tan(x) -1}$$

Isn't this answer (B)?

Correct...

My question is 'would a student be wrong in attempting to use separation of variables?' ...does he/she get method marks? cheers...

 

[chwala]

Your second equation above is equivalent to ##y = e^{\tan(x)}\cdot e^k## or ##y = Ce^{\tan(x)}##, where ##C= e^k##. Simply substitute ##\pi/4## for x and -1 for y to solve for C.

Won't we have ##e^k=-1##?

 

[Mark44]

Won't we have ##e^k=-1##?

No. Solve for C in this equation: ##-1 = Ce^1##

 

[Mark44]

My question is 'would a student be wrong in attempting to use separation of variables?' ...does he/she get method marks?

I don't see why the work would be marked as wrong. The question asks only which of the four listed equation is the solution and makes no requirement of which method to use.

 

[chwala]

Thanks @Mark44 ...my interest was on checking if we could use separation of variables. You have clearly shown that to me. Cheers!