Can you guys help me do this question? y'=2x-3y+1, y(1)=5, y(1.5)=?

[teyyyyyi]

Hi, I need someone to help me with the question above, I have currently solved it using the linear ordinary differential equations and I need to find 2 other new methods to find the EXACT value of it, not approximation value. Hence, I am not allowed to use RK-4, Euler's, and other approximating methods. Thank you!

 

[HallsofIvy]

This is strange! Isn't one method enough as long as it gives the correct answer? Yes, this is a linear equation of the form $\displaystyle \frac{dy}{dx}+ 3y= 2x+1$. The "integrating factor" is $\displaystyle e^{\int 3dx}= e^{3x}$. Multiplying both sides by that gives $\displaystyle e^{3x}\frac{dy}{dx}+ 3e^{3x}y= \frac{de^{3x}y}{dx}= (2x+ 1)e^{3x}$.
To integrate the right side use "integration by parts", letting $\displaystyle u= 2x+ 1$, $\displaystyle dv= e^{3x}$. Then $\displaystyle du= 2dx$ and $\displaystyle v= \frac{1}{3}x$.
So $\displaystyle \int (2x+ 1)e^{3x}dx= \frac{1}{3}(2x+1)e^{3x}- \frac{2}{3}\int e^{3x}dx= \frac{1}{3}(2x+ 1)e^{3x}- \frac{2}{9}e^{3x}+C= \frac{2}{3}xe^{3x}+ \frac{1}{9}e^{3x}+ C$. So $\displaystyle e^{3x}y= \frac{2}{3}xe^{3x}+ \frac{1}{9}e^{3x}+ C$ so $\displaystyle y= \frac{2}{3}x+ \frac{1}{9}+ Ce^{-3x}$.

Another method is separating into the "associated homogeneous equation" and the entire equation. The "associated homogeneous equation" is $\displaystyle \frac{dy}{dx}=-3y$ so $\displaystyle \frac{dy}{y}= -3dx$. Integrating that gives $\displaystyle ln(y)= -3x+ C$. Taking the exponential of both sides $\displaystyle y= e^{-3x+ C}= C'e^{-3x}$ where $\displaystyle C'= e^C$. Now we just need a single function that satisfies the entire equation. Since the "no-homogeneous" part is $\displaystyle 2x+1$ and I know that the derivative just lowers the degree of a polynomial, I look for a solution of the form $\displaystyle y= ax^2+ bx+ c$. Then $\displaystyle y'= 2ax+ b$ so the equation is $\displaystyle 2ax+ b+ 3ax^2+ 3bx+ 3c= 2x+ 1$. $\displaystyle 3ax^2+ (3b+ 2a)x+ (b+ 3c)= 2x+1$. Since this is to be true for all x the coefficients must be the same: 3a= 0, 3b+ 2a= 2 and b+ 3c= 1. That is, a= 0, b= 2/3, and c= 1/9.

The solution is $\displaystyle y= Ce{-3x}+ (2/3)x+ 1/9$ as before.

I'm not sure I would consider it a "different method" but we can also use the method used for higher order linear differential equations. Since the "associated homogeneous equation" is $\displaystyle \frac{dy}{dx}+ 3y= 0$, the "characteristic equation" is $\displaystyle r+3= 0$. $\displaystyle r= -3$ so the general solution to the associated homogeneous equation is $\displaystyle y= Ce^{-3x}$ and then finish as above.

 

[teyyyyyi]

This is strange! Isn't one method enough as long as it gives the correct answer? Yes, this is a linear equation of the form $\displaystyle \frac{dy}{dx}+ 3y= 2x+1$. The "integrating factor" is $\displaystyle e^{\int 3dx}= e^{3x}$. Multiplying both sides by that gives $\displaystyle e^{3x}\frac{dy}{dx}+ 3e^{3x}y= \frac{de^{3x}y}{dx}= (2x+ 1)e^{3x}$.
To integrate the right side use "integration by parts", letting $\displaystyle u= 2x+ 1$, $\displaystyle dv= e^{3x}$. Then $\displaystyle du= 2dx$ and $\displaystyle v= \frac{1}{3}x$.
So $\displaystyle \int (2x+ 1)e^{3x}dx= \frac{1}{3}(2x+1)e^{3x}- \frac{2}{3}\int e^{3x}dx= \frac{1}{3}(2x+ 1)e^{3x}- \frac{2}{9}e^{3x}+C= \frac{2}{3}xe^{3x}+ \frac{1}{9}e^{3x}+ C$. So $\displaystyle e^{3x}y= \frac{2}{3}xe^{3x}+ \frac{1}{9}e^{3x}+ C$ so $\displaystyle y= \frac{2}{3}x+ \frac{1}{9}+ Ce^{-3x}$.

Another method is separating into the "associated homogeneous equation" and the entire equation. The "associated homogeneous equation" is $\displaystyle \frac{dy}{dx}=-3y$ so $\displaystyle \frac{dy}{y}= -3dx$. Integrating that gives $\displaystyle ln(y)= -3x+ C$. Taking the exponential of both sides $\displaystyle y= e^{-3x+ C}= C'e^{-3x}$ where $\displaystyle C'= e^C$. Now we just need a single function that satisfies the entire equation. Since the "no-homogeneous" part is $\displaystyle 2x+1$ and I know that the derivative just lowers the degree of a polynomial, I look for a solution of the form $\displaystyle y= ax^2+ bx+ c$. Then $\displaystyle y'= 2ax+ b$ so the equation is $\displaystyle 2ax+ b+ 3ax^2+ 3bx+ 3c= 2x+ 1$. $\displaystyle 3ax^2+ (3b+ 2a)x+ (b+ 3c)= 2x+1$. Since this is to be true for all x the coefficients must be the same: 3a= 0, 3b+ 2a= 2 and b+ 3c= 1. That is, a= 0, b= 2/3, and c= 1/9.

The solution is $\displaystyle y= Ce{-3x}+ (2/3)x+ 1/9$ as before.

I'm not sure I would consider it a "different method" but we can also use the method used for higher order linear differential equations. Since the "associated homogeneous equation" is $\displaystyle \frac{dy}{dx}+ 3y= 0$, the "characteristic equation" is $\displaystyle r+3= 0$. $\displaystyle r= -3$ so the general solution to the associated homogeneous equation is $\displaystyle y= Ce^{-3x}$ and then finish as above.

thank you so much for your response! I appreciate it a lot ??. I hope this is not too much to ask, can you help me figure out other methods to use for y'=e^-y, y(0)=0, y(0.5)=? I have found the answer using separation of variables and I can't seem to find a new way, I tried using Laplace but it didn't work. Thank you before!

 

[teyyyyyi]

Hi, I need someone to help me with the question above, I have currently solved it using the separation of variables and I need to find 2 other new methods to find the EXACT value of it, not approximation value. Hence, I am not allowed to use RK-4, Euler's, and other approximating methods. Thank you!

 

[Deleted member 4993]

Hi, I need someone to help me with the question above, I have currently solved it using the separation of variables and I need to find 2 other new methods to find the EXACT value of it, not approximation value. Hence, I am not allowed to use RK-4, Euler's, and other approximating methods. Thank you!

Please share your work using "separation of variables".

Did you follow the methods pointed out in response #2?

 

[teyyyyyi]

Please share your work using "separation of variables".

Did you follow the methods pointed out in response #2?

Oh, I am sorry I was referring to another question for the "separation of variables", and yes I did the methods pointed out in response #2.

 

[Deleted member 4993]

Oh, I am sorry I was referring to another question for the "separation of variables", and yes I did the methods pointed out in response #2.

So what are the methods that you did try to solve the given DE?

Please share your work.

 

[teyyyyyi]

So what are the methods that you did try to solve the given DE?

Please share your work.

This is my work using the linear ordinary differential equations.