Solving Initial-value Problem: Different "C" Values Explained

[Astro]

Wondering why getting different values of "C" depending on how I solve the question. Not sure the values are different. Thanks.

1. Homework Statement
Solve the initial value problem $cos(x)Ln(y) \frac{dy} {dx} =ysin(x)$, y>0, y(0)=e².

Homework Equations

N/A.

The Attempt at a Solution

$∫ \frac{Ln(y)} {y} dy=∫ \frac{sin x} {cos x} dx$

u=Ln(y)
$du=\frac{1} {y}$ dy
dy=y du​

$∫ \frac{u} {y} y du=∫ tan(x) dx$
$\frac{1} {2}u^2+C_1=Ln|sec(x)| + C_2$
$\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$ ,Since $C=C_2-C_1$
$Ln(y)=± \sqrt{2 Ln|sec(x)|+C}$ Equation #1
$y=e^{± √(2Ln|sec(x)|+C)}$
--------------
$e^2=e^{+√(2Ln|sec(x)|+C)}$ Since y(0)=e². Also, exponent cannot be -ve since LS.=RS.
$e^2=e^{+√(2(0)+C)}$
$2=√C$
$C=4$

Therefore $y=e^{± √(2Ln|sec(x)|+4)}$ <--- ANS #1

However, if we try solving for C using eqⁿ #1:

$\frac{1}{2}Ln^2(y)=Ln|sec(x)|+C$
$\frac{1}{2}Ln^2(e^2)=Ln|sec(0)|+C$
$\frac{1}{2}(2^2)Ln^2(e)=Ln(1)+C$
$2(1)=0+C$
Therefore, $y=e^{± √(2Ln|sec(x)|+2)}$ <----ANS #2.

However, obviously ANS #1 ≠ ANS #2. Can someone please help me understand why and what I did wrong?
Thanks.

 

[Samy_A]

Wondering why getting different values of "C" depending on how I solve the question. Not sure the values are different. Thanks.

1. Homework Statement
Solve the initial value problem $cos(x)Ln(y) \frac{dy} {dx} =ysin(x)$, y>0, y(0)=e².

Homework Equations

N/A.

The Attempt at a Solution

$∫ \frac{Ln(y)} {y} dy=∫ \frac{sin x} {cos x} dx$

u=Ln(y)
$du=\frac{1} {y}$ dy
dy=y du​

$∫ \frac{u} {y} y du=∫ tan(x) dx$
$\frac{1} {2}u^2+C_1=Ln|sec(x)| + C_2$
$\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$ ,Since $C=C_2-C_1$
$Ln(y)=± \sqrt{2 Ln|sec(x)|+C}$ Equation #1
$y=e^{± √(2Ln|sec(x)|+C)}$
--------------
$e^2=e^{+√(2Ln|sec(x)|+C)}$ Since y(0)=e². Also, exponent cannot be -ve since LS.=RS.
$e^2=e^{+√(2(0)+C)}$
$2=√C$
$C=4$

Therefore $y=e^{± √(2Ln|sec(x)|+4)}$ <--- ANS #1

However, if we try solving for C using eqⁿ #1:

$\frac{1}{2}Ln^2(y)=Ln|sec(x)|+C$
$\frac{1}{2}Ln^2(e^2)=Ln|sec(0)|+C$
$\frac{1}{2}(2^2)Ln^2(e)=Ln(1)+C$
$2(1)=0+C$
Therefore, $y=e^{± √(2Ln|sec(x)|+2)}$ <----ANS #2.

However, obviously ANS #1 ≠ ANS #2. Can someone please help me understand why and what I did wrong?
Thanks.

You don't use eqⁿ #1 in the second part, but the equation immediately above it:
$\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$
Notice that $C_3$ is not equal to your final $C$.

 

[mfb]

$\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$ ,Since $C=C_2-C_1$
$Ln(y)=± \sqrt{2 Ln|sec(x)|+C}$ Equation #1

The two "C" here differ by a factor of 2.
Approach 1 is correct, approach 2 gives you C₃ = 1/2 C = 2 which leads to C=4 as well.

 

[Astro]

I'm not sure I follow.

To solve for the correct answer for "C" (which is C=4 according to the textbook), all I did was rearrange $\frac{1}{2}Ln^2(y)=Ln|sec(x)|+C$ to solve for "y" and then subbing in the initial values x=0, and y=e^{^2} gives you $e^2=e^{\sqrt{C}}$ from which you get C=4. But, when I sub in the initial values directly into $\frac{1}{2}Ln^2(y)=Ln|sec(x)|+C$ and rearrange for C I get a different answer. It's unclear to me why I'm getting different answers because in both cases the variable C is the same. Solving for C two different ways when you're only rearranging the equation differently to find C should give you the same answer if you apply the rules of logarithms and exponents correctly. The fact that I'm not getting the same value of C means that I'm not understanding some basic concept of math or that I'm making a mistake when I'm rearranging and solving for C.

 

[ehild]

$\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$ ,Since $C=C_2-C_1$
$Ln(y)=± \sqrt{2 Ln|sec(x)|+C}$ Equation #1
$y=e^{± √(2Ln|sec(x)|+C)}$
.

You forgot to multiply C2-C1 by 2.

 

[Samy_A]

I'm not sure I follow.

To solve for the correct answer for "C" (which is C=4 according to the textbook), all I did was rearrange $\frac{1}{2}Ln^2(y)=Ln|sec(x)|+C$ to solve for "y" and then subbing in the initial values x=0, and y=e^{^2} gives you $e^2=e^{\sqrt{C}}$ from which you get C=4. But, when I sub in the initial values directly into $\frac{1}{2}Ln^2(y)=Ln|sec(x)|+C$ and rearrange for C I get a different answer. It's unclear to me why I'm getting different answers because in both cases the variable C is the same. Solving for C two different ways when you're only rearranging the equation differently to find C should give you the same answer if you apply the rules of logarithms and exponents correctly. The fact that I'm not getting the same value of C means that I'm not understanding some basic concept of math or that I'm making a mistake when I'm rearranging and solving for C.

(bolding mine)
What I bolded is wrong.

$\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$ ,Since $C=C_2-C_1$
$Ln(y)=± \sqrt{2 Ln|sec(x)|+C}$ Equation #1
$y=e^{± √(2Ln|sec(x)|+C)}$

When you solve for $C$ using $y=e^{± √(2Ln|sec(x)|+C)}$, you find $C=4$. That is correct.
When you solve for $C_3$ using $\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$, you find $C_3=2$. That is correct too.
But, as mfb, ehild and myself have mentionned, the constants in these two equations are not equal. You confused yourself by renaming $C_3$ to $C$ when you tried to find the constant using $\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$.
Let's rewrite you equations keeping $C_3$:
$\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3$
$Ln(y)=± \sqrt{2 Ln|sec(x)|+2C_3}$
$y=e^{± √(2Ln|sec(x)|+2C_3)}$.

You see? Your final $C=2C_3$, that's why you find 4 using one equation, and 2 using the other.