Engineering Help - Direct Integration - Particular Rule

[LukeGernon]

So I have cracked 5 a and B but boy I don't even know where to begin with question 6. Im looking for some help hopefully theres some similar worked examples out there I can take a look at if someone could point me in the right direction.

 

[Deleted member 4993]

So I have cracked 5 a and B but boy I don't even know where to begin with question 6. Im looking for some help hopefully theres some similar worked examples out there I can take a look at if someone could point me in the right direction.View attachment 22361

Hint: First determine the solution of the homogeneous DE. If one of the solutions is "cos(2x)" - then we have a "resonance" situation.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

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Please share your work/thoughts about this problem.

 

[LukeGernon]

Here is what I’ve got to using the formula given in the question though I feel like you’re meant to re arrange it before you begin but I wasn’t able to find a solution that looked correct.

 

[LukeGernon]

Re arranging formula attempt

 

[HallsofIvy]

Personally, I would consider 6 easier than 5!

Problem 6 is a "linear equation with constant coefficients" and there is a standard method for solving it. The "associated homogeneous equation" is \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= 0\). It's "characteristic equation" is \(\displaystyle r^2+ 3r+ 2=(r+ 1)(r+ 2)= 0\) so its "characteristic roots" are -1 and -2. That tells us that the general solution to the associated homogeneous equation is \(\displaystyle C_e^{-x}+ C_2e^{-2x}\).

Now, to find the general solution to the entire equation, we just need to add any function that satisfies the entire equation. And recalling that the derivatives of sine and cosine are always again sine and cosine, I try a solution of the form \(\displaystyle y= Asin(2x)+ Bcos(2x)\) where I want to determine numbers A and B.

With \(\displaystyle y= Asin(2x)+ Bcos(2x)\), \(\displaystyle \frac{dy}{dx}= 2Acos(2x)- 2Bsin(2x)\) and \(\displaystyle \frac{d^2y}{dx^2}= -4A sin(2x)- 4B cos(2x)\).

So the equation becomes \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= -4A sin(2x)- 4B cos(2x)+ 6A cos(2x)- 6B sin(2x)+ 2A sin(2x)+ 2B cos(2x)= (-4A- 6B+ 2A) sin(2x)+ (-4B+ 6A+ 2B) cos(2x)= (-2A- 6B) sin(2x)+ (-2B+ 6A) cos(2x)= 10 cos(2x)\).

Since this is to be true for all x we must have -2A- 6B= 0 and -2B+ 6A= 10. Multiply the first equation by 2 to get -4A- 12B= 0. Multiply the second equation by 6 to get -12B+ 36A= 60. Subtract the first of those from the second to eliminate B and get 40A= 60 so A= 6/4= 3/2. The -4A- 12B= -6- 12B= 0. -12B= 6 so B= -6/12= -1/2.

The general solution to the entire equation is \(\displaystyle y(x)= C_1e^{-x}+ C_2e^{-2x}+ \frac{3}{2}sin(2x)- \frac{1}{2} cos(2x)\). The last thing to do, of course, is to determine \(\displaystyle C_1\) and \(\displaystyle C_2\) so that y(0)= 1 and y'(0)= 0.
\(\displaystyle y(0)= C_1+ C_2+ \frac{1}{2}= 1\).
\(\displaystyle y'(x)= -C_1e^{-x}- 2C_2e^{-x}+ 3 cos(2x)+ sin(2x)\) so
\(\displaystyle y'(0)= -C_1- 2C_2+ 3= 0\).

 

[Deleted member 4993]

Personally, I would consider 6 easier than 5!

Problem 6 is a "linear equation with constant coefficients" and there is a standard method for solving it. The "associated homogeneous equation" is \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= 0\). It's "characteristic equation" is \(\displaystyle r^2+ 3r+ 2=(r+ 1)(r+ 2)= 0\) so its "characteristic roots" are -1 and -2. That tells us that the general solution to the associated homogeneous equation is \(\displaystyle C_e^{-x}+ C_2e^{-2x}\).

Now, to find the general solution to the entire equation, we just need to add any function that satisfies the entire equation. And recalling that the derivatives of sine and cosine are always again sine and cosine, I try a solution of the form \(\displaystyle y= Asin(2x)+ Bcos(2x)\) where I want to determine numbers A and B.

With \(\displaystyle y= Asin(2x)+ Bcos(2x)\), \(\displaystyle \frac{dy}{dx}= 2Acos(2x)- 2Bsin(2x)\) and \(\displaystyle \frac{d^2y}{dx^2}= -4A sin(2x)- 4B cos(2x)\).

So the equation becomes \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= -4A sin(2x)- 4B cos(2x)+ 6A cos(2x)- 6B sin(2x)+ 2A sin(2x)+ 2B cos(2x)= (-4A- 6B+ 2A) sin(2x)+ (-4B+ 6A+ 2B) cos(2x)= (-2A- 6B) sin(2x)+ (-2B+ 6A) cos(2x)= 10 cos(2x)\).

Since this is to be true for all x we must have -2A- 6B= 0 and -2B+ 6A= 10. Multiply the first equation by 2 to get -4A- 12B= 0. Multiply the second equation by 6 to get -12B+ 36A= 60. Subtract the first of those from the second to eliminate B and get 40A= 60 so A= 6/4= 3/2. The -4A- 12B= -6- 12B= 0. -12B= 6 so B= -6/12= -1/2.

The general solution to the entire equation is \(\displaystyle y(x)= C_1e^{-x}+ C_2e^{-2x}+ \frac{3}{2}sin(2x)- \frac{1}{2} cos(2x)\). The last thing to do, of course, is to determine \(\displaystyle C_1\) and \(\displaystyle C_2\) so that y(0)= 1 and y'(0)= 0.
\(\displaystyle y(0)= C_1+ C_2+ \frac{1}{2}= 1\).
\(\displaystyle y'(x)= -C_1e^{-x}- 2C_2e^{-x}+ 3 cos(2x)+ sin(2x)\) so
\(\displaystyle y'(0)= -C_1- 2C_2+ 3= 0\).

I fully agree! I was almost tempted to show work for solved #5.

 

[LukeGernon]

Personally, I would consider 6 easier than 5!

Problem 6 is a "linear equation with constant coefficients" and there is a standard method for solving it. The "associated homogeneous equation" is \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= 0\). It's "characteristic equation" is \(\displaystyle r^2+ 3r+ 2=(r+ 1)(r+ 2)= 0\) so its "characteristic roots" are -1 and -2. That tells us that the general solution to the associated homogeneous equation is \(\displaystyle C_e^{-x}+ C_2e^{-2x}\).

Now, to find the general solution to the entire equation, we just need to add any function that satisfies the entire equation. And recalling that the derivatives of sine and cosine are always again sine and cosine, I try a solution of the form \(\displaystyle y= Asin(2x)+ Bcos(2x)\) where I want to determine numbers A and B.

With \(\displaystyle y= Asin(2x)+ Bcos(2x)\), \(\displaystyle \frac{dy}{dx}= 2Acos(2x)- 2Bsin(2x)\) and \(\displaystyle \frac{d^2y}{dx^2}= -4A sin(2x)- 4B cos(2x)\).

So the equation becomes \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= -4A sin(2x)- 4B cos(2x)+ 6A cos(2x)- 6B sin(2x)+ 2A sin(2x)+ 2B cos(2x)= (-4A- 6B+ 2A) sin(2x)+ (-4B+ 6A+ 2B) cos(2x)= (-2A- 6B) sin(2x)+ (-2B+ 6A) cos(2x)= 10 cos(2x)\).

Since this is to be true for all x we must have -2A- 6B= 0 and -2B+ 6A= 10. Multiply the first equation by 2 to get -4A- 12B= 0. Multiply the second equation by 6 to get -12B+ 36A= 60. Subtract the first of those from the second to eliminate B and get 40A= 60 so A= 6/4= 3/2. The -4A- 12B= -6- 12B= 0. -12B= 6 so B= -6/12= -1/2.

The general solution to the entire equation is \(\displaystyle y(x)= C_1e^{-x}+ C_2e^{-2x}+ \frac{3}{2}sin(2x)- \frac{1}{2} cos(2x)\). The last thing to do, of course, is to determine \(\displaystyle C_1\) and \(\displaystyle C_2\) so that y(0)= 1 and y'(0)= 0.
\(\displaystyle y(0)= C_1+ C_2+ \frac{1}{2}= 1\).
\(\displaystyle y'(x)= -C_1e^{-x}- 2C_2e^{-x}+ 3 cos(2x)+ sin(2x)\) so
\(\displaystyle y'(0)= -C_1- 2C_2+ 3= 0\).

Thankyou very much, seems I was missing quite some bits. Appreciate your help.

 

[LukeGernon]

Personally, I would consider 6 easier than 5!

Problem 6 is a "linear equation with constant coefficients" and there is a standard method for solving it. The "associated homogeneous equation" is \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= 0\). It's "characteristic equation" is \(\displaystyle r^2+ 3r+ 2=(r+ 1)(r+ 2)= 0\) so its "characteristic roots" are -1 and -2. That tells us that the general solution to the associated homogeneous equation is \(\displaystyle C_e^{-x}+ C_2e^{-2x}\).

Now, to find the general solution to the entire equation, we just need to add any function that satisfies the entire equation. And recalling that the derivatives of sine and cosine are always again sine and cosine, I try a solution of the form \(\displaystyle y= Asin(2x)+ Bcos(2x)\) where I want to determine numbers A and B.

With \(\displaystyle y= Asin(2x)+ Bcos(2x)\), \(\displaystyle \frac{dy}{dx}= 2Acos(2x)- 2Bsin(2x)\) and \(\displaystyle \frac{d^2y}{dx^2}= -4A sin(2x)- 4B cos(2x)\).

So the equation becomes \(\displaystyle \frac{d^2y}{dx^2}+ 3\frac{dy}{dx}+ 2y= -4A sin(2x)- 4B cos(2x)+ 6A cos(2x)- 6B sin(2x)+ 2A sin(2x)+ 2B cos(2x)= (-4A- 6B+ 2A) sin(2x)+ (-4B+ 6A+ 2B) cos(2x)= (-2A- 6B) sin(2x)+ (-2B+ 6A) cos(2x)= 10 cos(2x)\).

Since this is to be true for all x we must have -2A- 6B= 0 and -2B+ 6A= 10. Multiply the first equation by 2 to get -4A- 12B= 0. Multiply the second equation by 6 to get -12B+ 36A= 60. Subtract the first of those from the second to eliminate B and get 40A= 60 so A= 6/4= 3/2. The -4A- 12B= -6- 12B= 0. -12B= 6 so B= -6/12= -1/2.

The general solution to the entire equation is \(\displaystyle y(x)= C_1e^{-x}+ C_2e^{-2x}+ \frac{3}{2}sin(2x)- \frac{1}{2} cos(2x)\). The last thing to do, of course, is to determine \(\displaystyle C_1\) and \(\displaystyle C_2\) so that y(0)= 1 and y'(0)= 0.
\(\displaystyle y(0)= C_1+ C_2+ \frac{1}{2}= 1\).
\(\displaystyle y'(x)= -C_1e^{-x}- 2C_2e^{-x}+ 3 cos(2x)+ sin(2x)\) so
\(\displaystyle y'(0)= -C_1- 2C_2+ 3= 0\).

so how do we find the constants they’re still unknowns right?