You CAN Ace Differential Equations

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17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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General Form: \(y'' + p(t)y' + q(t)y = g(t)\)

classification: second-order, linear, constant coefficients, inhomogeneous → \(g(t) \neq 0\)

The idea of this technique is to use \(g(t)\) to guess what the form of the solution looks like.

For certain functions, the solutions will look a lot like the given \(g(t)\). However, this works only for certain standard forms of \(g(t)\). This technique for handling second order inhomogeneous linear differential equations is great when \(g(t)\) is in a standard form. Here is a list of some of the standard forms that this technique will work with.

1. polynomials (including constants)

2. exponentials in the form \(e^{at}\)

3. trig functions \(\sin(bt)\) and \(\cos(ct)\)

If \(g(t)\) is not one of the standard forms above, the method of variation of parameters, which is a bit more involved, needs to be used.

General Inhomogeneous Solutions

The general solution to an inhomogeneous differential equation is made up of two parts, the homogeneous solution plus what many mathematicians call the particular solution. The idea is that you can add zero to anything you don't change the problem. So, you can look at the differential equation
\(\displaystyle{ y'' + p(t)y' + q(t)y = g(t) = 0 + g(t) }\)
So we find the homogeneous solution, \(y_H(t)\) which solves \(y'' + p(t)y' + q(t)y = 0\) and add it to the inhomogeneous solution or particular solution \(y_P(t)\) which solves \(y'' + p(t)y' + q(t)y = g(t)\) to get the general solution \(y(t) = y_H(t) + y_P(t)\).

Before we start discussing how to get solutions to inhomogeneous equations using undetermined coefficients, let's watch a video talking about inhomogeneous equations is general.

MIT OCW - solutions [mins-secs]

video by MIT OCW

Method of Undetermined Coefficients

The discussion on this page assumes you know how to get the homogeneous solution or it is given to you. We will work to get the inhomogeneous solution. You need to know what the homogeneous solution is first before using this technique. The reason is that sometimes you will need to adjust your 'guess' based on the form of the homogeneous solution.

Let\'s start out with a video. This video explains very well, this idea of a general solution built from the homogeneous and particular solutions, as well as the method of undetermined coefficients. It is a bit longer than others you will find on this site but this one is especially worth taking the time to watch. He does several related examples that build on one another and will help you understand what to do. At the very end, he also explains why this technique works. So this is one video you don\'t want to skip.

Dr Chris Tisdell - general solution [mins-secs]

video by Dr Chris Tisdell

If you want another quick overview of this method, here is a quick video clip that might help cement this technique in your mind.

PatrickJMT - quick overview [1mins-33secs]

video by PatrickJMT

Okay, let's look at each of the standard forms individually and see what your initial 'guesses' look like, as well as variations that are required to avoid duplication with the homogeneous solution. In the following tables, we are talking about the roots to the characteristic polynomial that you used to find the homogeneous solution.

Standard Form 1 - - Polynomials

\(g(t) = a_2t^2 + a_1t + a_0\)

basic form

\(y_P = At^2+Bt+C\)

variations

\(r=0\) is a single root

\(y_P = ( At^2+Bt+C)t \)

\(r=0\) is a double root

\(y_P = ( At^2+Bt+C)t^2 \)

Note - - In the table above, we showed a second order polynomial. In general, your basic form polynomial needs to be of the same order as your \(g(t)\) polynomial. For example, if \(g(t) = t^4\), your basic polynomial is \(y_P = At^4 + Bt^3 + Ct^2 + Dt + E \), then you would adjust based on the roots to the homogeneous solution, if necessary.

Standard Form 2 - - Exponentials

\(g(t) = e^{ct}\)

basic form

\(y_P = Ae^{ct}\)

variations

\(r=c\) is a single root

\(y_P = Ate^{ct} \)

\(r=c\) is a double root

\(y_P = At^2e^{ct} \)

Standard Form 3 - - Trig Functions

\(g(t) = a_0\sin(kt)\) or \(g(t) = a_0\cos(kt)\) or both

basic form

\(y_P = A\sin(kt)+B\cos(kt)\)

In each case, the function that you build based on \(g(t)\) must be different than any factor in the homogeneous solution, i.e. it must not be a constant multiple of any other factor. You may need to multiply by \(t\) or \(t^2\) to accomplish this, but, as you saw in the first video above, doing so will resolve any conflicts.

After working some practice problems, you will be ready for the next topic, variation of parameters, which is a technique that handles a larger range of problems when \(g(t)\) is not one of the above standard forms.

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, find the general solution to each differential equation using the method of undetermined coefficients. If initial conditions are given, find the particular solution also. Give your answers in exact terms and completely factored.

Basic Problems

\( y'' + 4y = e^{2x} \)

Problem Statement

\( y'' + 4y = e^{2x} \)

Final Answer

\( y = c_1 \cos(2x) + c_2 \sin(2x) + (1/8)e^{2x} \)

Problem Statement

\( y'' + 4y = e^{2x} \)

Solution

First, we need to solve the homogeneous equation \( y'' + 4y = 0 \).
\( r^2+4 = 0 ~~~ \to ~~~ r^2 = -4 ~~~ \to ~~~ r = \pm 2i \)
\( y_h = c_1 \cos(2x) + c_2 \sin(2x) \)
So the form of the non-homogeneous solution is \( y_p(x) = Ae^{2x} \)
\( y'_p(x) = 2Ae^{2x} ~~~~~~ y''_p(x) = 4Ae^{2x} \)
\(\begin{array}{rcl} y''_p + 4y_p & = & e^{2x} \\ 4Ae^{2x} + 4(Ae^{2x}) & = & e^{2x} \\ 8Ae^{2x} & = & e^{2x} \\ 8A & = & 1 \\ A & = & 1/8 \end{array}\)
Combining the two solutions gives us the final answer, i.e. \( y = y_h + y_p \).

Final Answer

\( y = c_1 \cos(2x) + c_2 \sin(2x) + (1/8)e^{2x} \)

close solution

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)

Problem Statement

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)

Final Answer

\(\displaystyle{ y(x) = \frac{7}{8}\cos(2x) + \frac{11}{8}\sin(2x) + \frac{1}{8}e^{2x} }\)

Problem Statement

\( y'' + 4y = e^{2x}; y(0)=1, y'(0)=3 \)

Solution

From the previous problem, we have the general solution \(\displaystyle{ y = c_1 \cos(2x) + c_2 \sin(2x) + \frac{1}{8} e^{2x}}\)
Using the initial conditions to find the constants \(c_1\) and \(c_2\), we have
\( y(0) = c_1 + 1/8 = 1 ~~~ \to ~~~ c_1 = 1-1/8 = 7/8 \)

\(\displaystyle{ y'(x) = -2c_1 \sin(2x) + 2c_2 \cos(2x) + \frac{1}{8}(2)e^{2x} }\)

\( y'(0) = 2c_2 + 1/4 = 3 ~~~ \to ~~~ 2c_2 = 3-1/4 = 11/4 ~~~ \to ~~~ c_2 = 11/8 \)

Final Answer

\(\displaystyle{ y(x) = \frac{7}{8}\cos(2x) + \frac{11}{8}\sin(2x) + \frac{1}{8}e^{2x} }\)

close solution

\( y''-4y = e^{2x} \)

Problem Statement

\( y''-4y = e^{2x} \)

Final Answer

\( y = c_1e^{2x} + c_2 e^{-2x} + (x/4)e^{2x} \)

Problem Statement

\( y''-4y = e^{2x} \)

Solution

First, we need to find the solution to the homogeneous equation, \( y'' - 4y = 0 \).

\( y'' - 4y = 0 \)

characteristic equation

\( r^2 - 4 = 0 \)

\( r^2 = 4 \)

\( r = \pm 2 \)

homogeneous solution

\( y_h(x) = c_1 e^{2x} + c_2 e^{-2x} \)

Now the form of the non-homogeneous (particular) solution is \( y_p(x) = Axe^{2x} \).

\( y_p(x) = Axe^{2x} \)

\( y'_p = Ax(2e^{2x}) + Ae^{2x} = (2Ax+A)e^{2x} \)

\( y''_p = (2Ax+A)(2e^{2x}) + e^{2x}(2A) = (4Ax+2A+2A)e^{2x} = (4Ax+4A)e^{2x} \)

plug these expressions into the original differential equation

\( y''_p - 4y_p = e^{2x} \)

\( (4Ax+4A)e^{2x} - 4(Axe^{2x}) = e^{2x} \)

\( 4Ae^{2x} = e^{2x} \)

\( 4A = 1 \)

\( A = 1/4 \)

particular solution

\( (x/4)e^{2x} \)

Combining the homogeneous and particular solutions gives us the complete solution, \( y = y_h + y_p \)

Final Answer

\( y = c_1e^{2x} + c_2 e^{-2x} + (x/4)e^{2x} \)

close solution

\( y'' - 2y' + y = (2x+1)e^x \)

Problem Statement

\( y'' - 2y' + y = (2x+1)e^x \)

Final Answer

\( \displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }\)

Problem Statement

\( y'' - 2y' + y = (2x+1)e^x \)

Solution

As before, we need to find the solutions to the homogeneous equation.

\( y''-2y'+y = 0 \)

characteristic equation

\( r^2 - 2r + 1 = 0 \)

\( (r-1)^2 = 0 \)

\( r = 1 \) multiplicity 2

homogeneous solution

\( y_h = c_1 e^x + c_2 x e^x \)

So the form of the non-homogeneous (particular) solution is \( y_p = (Ax+B)e^x (x^2) = (Ax^3+Bx^2)e^x \)

\( y_p = (Ax^3+Bx^2)e^x \)

\( y'_p = (Ax^3+Bx^2)e^x + e^x(3Ax^2+2Bx) \)

\( y'_p = (Ax^3 + 3Ax^2 + Bx^2 + 2Bx)e^x \)

\( y''_p = (Ax^3 + 3Ax^2 +Bx^2+2Bx)e^x + e^x(3Ax^2+6Ax+2Bx+2B) \)

\( y''_p = (Ax^3 + 6Ax^2 + Bx^2 +4Bx +2B +6Ax) e^x \)

Now plug these expressions into the original differential equation.

\( y''_p - 2y'_p + y_p \)

\( (Ax^3 + 6Ax^2 + Bx^2 +4Bx +2B +6Ax) e^x \)

\( -2(Ax^3 + 3Ax^2 +Bx^2+2Bx)e^x + (Ax^3 + Bx^2)e^x \)

\( (6Ax + 2B)e^x = (2x+1)e^x \)

Equating coefficients in the last equation yields

\( 2B = 1 ~~~ \to ~~~ B = 1/2 \)

\( 6A = 2 ~~~ \to ~~~ A = 1/3 \)

particular solution

\(\displaystyle{ y_p = \frac{x^3}{3} + \frac{x^2}{2} }\)

The final answer combines the results from the homogeneous and non-homgeneous solutions, \( y = y_h + y_p \).

Final Answer

\( \displaystyle{ y = \left[ \frac{x^3}{3} + \frac{x^2}{2} + c_2 x + c_1 \right] e^x }\)

close solution

\( y'' - 5y' + 6y = 4e^t \)

Problem Statement

\( y'' - 5y' + 6y = 4e^t \)

Final Answer

\( y(t) = 2e^t + c_1e^{2t} + c_2e^{3t} \)

Problem Statement

\( y'' - 5y' + 6y = 4e^t \)

Solution

First, we solve the homogeneous equation \( y'' - 5y' + 6y = 0 \)

The characteristic equation is

\( r^2 - 5r + 6 = 0 \)

\( (r-3)(r-2) = 0 \)

So our homogeneous solution is

\( y_H(t) = c_1e^{2t} + c_2e^{3t} \)

Now we will assume our particular solution is of the form \(y_p = Ae^t\).

\( y_p(t) = Ae^t \)

\( y'_p(t) = Ae^t \)

\( y''_p(t) = Ae^t \)

\( Ae^t - 5Ae^t + 6Ae^t = 4e^t \)

\( 2Ae^t = 4e^t \)

\( A = 2 \)

Final Answer

\( y(t) = 2e^t + c_1e^{2t} + c_2e^{3t} \)

close solution

\( \dot{x} + 3x = t^2 + t \)

Problem Statement

\( \dot{x} + 3x = t^2 + t \)

Final Answer

\(\displaystyle{ x(t) = Ae^{-3t} + \frac{1}{3}t^2 + \frac{1}{9}t - \frac{1}{27} }\)

Problem Statement

\( \dot{x} + 3x = t^2 + t \)

Solution

In the video, he does not determine the homogeneous solution, which is recommended on this website and by most instructors. Here are the steps to do that.

\( \dot{x} + 3x \) tells us that our characteristic equation is \(r + 3 = 0 \to r = -3\). So our homogeneous solution is \(\displaystyle{x_H(t) = Ae^{-3t} }\)

He determines the particular solution. So the final solution is \(x(t) = x_H(t) + x_P(t)\)

610 solution video

video by MIT OCW

Final Answer

\(\displaystyle{ x(t) = Ae^{-3t} + \frac{1}{3}t^2 + \frac{1}{9}t - \frac{1}{27} }\)

close solution

\( \ddot{x} + \dot{x} = t^4 \)

Problem Statement

\( \ddot{x} + \dot{x} = t^4 \)

Final Answer

\(\displaystyle{ x(t) = A + Be^{-t} + \frac{1}{5}t^5 - t^4 + 4t^3 - 12t^2 + 12t }\)

Problem Statement

\( \ddot{x} + \dot{x} = t^4 \)

Solution

611 solution video

video by MIT OCW

Final Answer

\(\displaystyle{ x(t) = A + Be^{-t} + \frac{1}{5}t^5 - t^4 + 4t^3 - 12t^2 + 12t }\)

close solution

\( y'' + 3y' + 2y = x^2 \)

Problem Statement

\( y'' + 3y' + 2y = x^2 \)

Final Answer

\(\displaystyle{ y(x) = c_1 e^{-x} + c_2e^{-2x} + \frac{1}{2}x^2 - \frac{3}{2}x + \frac{7}{4} }\)

Problem Statement

\( y'' + 3y' + 2y = x^2 \)

Solution

The solution to this problem appears in two videos. The first video solves for the homogeneous solution. The second video uses the undetermined coefficient technique to finish the problem.

612 solution video

video by PatrickJMT

612 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ y(x) = c_1 e^{-x} + c_2e^{-2x} + \frac{1}{2}x^2 - \frac{3}{2}x + \frac{7}{4} }\)

close solution

\( y'' - 5y' + 6y = 2x + 3 \)

Problem Statement

\( y'' - 5y' + 6y = 2x + 3 \)

Final Answer

\( y = Ae^{3x} + Be^{2x} + x/3 + 7/9 \)

Problem Statement

\( y'' - 5y' + 6y = 2x + 3 \)

Solution

613 solution video

video by Dr Chris Tisdell

Final Answer

\( y = Ae^{3x} + Be^{2x} + x/3 + 7/9 \)

close solution

\( y'' - 5y' + 6y = 12e^{5x} \)

Problem Statement

\( y'' - 5y' + 6y = 12e^{5x} \)

Final Answer

\( y = Ae^{3x} + Be^{2x} + 2e^{5x} \)

Problem Statement

\( y'' - 5y' + 6y = 12e^{5x} \)

Solution

614 solution video

video by Dr Chris Tisdell

Final Answer

\( y = Ae^{3x} + Be^{2x} + 2e^{5x} \)

close solution

\( y'' - 3y' - 4y = 3e^{2x} \)

Problem Statement

\( y'' - 3y' - 4y = 3e^{2x} \)

Final Answer

\(\displaystyle{ y = c_1e^{4x} + c_2e^{-x} - \frac{1}{2} e^{2x} }\)

Problem Statement

\( y'' - 3y' - 4y = 3e^{2x} \)

Solution

616 solution video

video by Khan Academy

Final Answer

\(\displaystyle{ y = c_1e^{4x} + c_2e^{-x} - \frac{1}{2} e^{2x} }\)

close solution

\( y'' + 3y' - 4y = 2e^x; y(0)=1, y'(0)=2\)

Problem Statement

\( y'' + 3y' - 4y = 2e^x; y(0)=1, y'(0)=2\)

Final Answer

\(\displaystyle{ y(x) = \frac{-3}{25}e^{-4x} + \frac{28}{25}e^x + \frac{2}{5}xe^x }\)

Problem Statement

\( y'' + 3y' - 4y = 2e^x; y(0)=1, y'(0)=2\)

Solution

624 solution video

video by Krista King Math

Final Answer

\(\displaystyle{ y(x) = \frac{-3}{25}e^{-4x} + \frac{28}{25}e^x + \frac{2}{5}xe^x }\)

close solution

Intermediate Problems

\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)

Problem Statement

\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)

Final Answer

\(\displaystyle{y(x)=e^{-x/2}\left[\frac{80}{17}\cos(x)-\right.}\) \(\displaystyle{\left.\frac{20}{17}\sin(x)\right]-}\) \(\displaystyle{\frac{80}{17}\cos(x)+}\) \(\displaystyle{\frac{20}{17}\sin(x)}\)
steady state occurs as \(x\to\infty\); so \(e^{-x/2}\to 0\) and \(\displaystyle{y_{ss}(x)=\frac{-80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)

Problem Statement

\( y'' + y' + 1.25y = 5\sin(x); y(0) = y'(0) = 0 \)

Solution

\(\begin{array}{rcl} r^2+r+1.25 & = & 0 \\ r^2+r+1/4 & = & -5/4+1/4 \\ (r+1/2)^2 & = & -1 \\ r+1/2 & = & \pm i \\ r & = & -1/2 \pm i \end{array}\)
\(y_h=e^{-x/2}[c_1\cos(x)+c_2\sin(x)]\)
\(\begin{array}{rcl} y_p & = & A\cos(x)+B\sin(x) \\ y'_p & = & -A\sin(x)+B\cos(x) \\ y''_p & = & -A\cos(x)-B\sin(x) \end{array}\)

\( \displaystyle{y''_p+y'_p+\frac{5}{4}y} \)

\( \displaystyle{[-A\cos(x)-B\sin(x)]+[-A\sin(x)+B\cos(x)]+\frac{5}{4}[A\cos(x)+B\sin(x)]} \)

\( [-A+B+5A/4]\cos(x)+[-B-A+5B/4]\sin(x)=5\sin(x) \)

Equating coefficients, we have two equations and two unknowns.

cosine term

   

sine term

\(B+A/4=0\)\(-A+B/4=5\)
\(4B+A=0\)\(-4A+B=20\)
\(B=20+4A\)
\(4(20+4A)+A=0\)
\(80+16A+A=0\)
\(17A=-80\)
\(A=-80/17 \)
\(B=20+4(-80/17)\)
\(B=(340-320)/17\)
\(B=20/17\)

So far, we have \(\displaystyle{y=e^{-x/2}[c_1\cos(x)+c_2\sin(x)]-\frac{80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)
Now we use the initial conditions to determine the constants \(c_1\) and \(c_2\).

\(y(0)=0\)

\(y(0)=1(c_1+0)-(80/17)(1)+0=0 ~~~ \to ~~~ c_1=80/17\)

\(y'(0)=0\)

\(y'=e^{-x/2}[-c_1\sin(x)+c_2\cos(x)]+(80/17)\sin(x)+(20/17)\cos(x)\)
\(y'(0)=1(0+c_2)+0+20/17=0 ~~~ \to ~~~ c_2=-20/17\)

Final Answer

\(\displaystyle{y(x)=e^{-x/2}\left[\frac{80}{17}\cos(x)-\right.}\) \(\displaystyle{\left.\frac{20}{17}\sin(x)\right]-}\) \(\displaystyle{\frac{80}{17}\cos(x)+}\) \(\displaystyle{\frac{20}{17}\sin(x)}\)
steady state occurs as \(x\to\infty\); so \(e^{-x/2}\to 0\) and \(\displaystyle{y_{ss}(x)=\frac{-80}{17}\cos(x)+\frac{20}{17}\sin(x)}\)

close solution

\( y^{(3)} -4y' = 4e^{2t} \)

Problem Statement

Use the method of undetermined coefficients to find the general solution to \( y^{(3)} -4y' = 4e^{2t} \)

Final Answer

\( \displaystyle{ y = y_h + y_p = A + Be^{2t} + Ce^{-2t} + \frac{te^{2t}}{2} }\)

Problem Statement

Use the method of undetermined coefficients to find the general solution to \( y^{(3)} -4y' = 4e^{2t} \)

Solution

First, we solve the homogeneous equation \( y^{(3)} -4y' = 0 \).
\(\begin{array}{rcl} r^3 - 4r & = & 0 \\ r(r^2-4) & = & 0 \\ r = 0 & & r^2 = 4 \to r = \pm 2 \end{array}\)
\( y_h = A + Be^{2t} + Ce^{-2t} \)
Now we need to find the particular solution associated with \( 4e^{2t} \).
Since we already have an \( e^{2t} \) factor in the homogeneous solution, we need to use \( y_p = Dte^{2t} \), which includes the extra factor of \( t \) to obtain an independent solution.

\( y'_p = Dt(2e^{2t}) + De^{2t} = (2Dt+D)e^{2t} \)

\( y''_p = (2Dt+D)e^{2t}(2) + (2D)e^{2t} = (4Dt+4D)e^{2t} \)

\( y^{(3)}_p = (4Dt+4D)e^{2t}(2) + e^{2t}(4D) = (8Dt+12D)e^{2t} \)

\( y^{(3)} -4y' = 4e^{2t} \)

\( (8Dt+12D)e^{2t} - 4(2DT+D)e^{2t} = 4e^{2t} \)

\( (12D - 4D)e^{2t} = 4e^{2t} \)

\( 8D = 4 \)

\( D = 1/2 \)

\( \displaystyle{ y_p = \frac{te^{2t}}{2} }\)

Combining the homogeneous and particular solutions gives us the general solution.

Final Answer

\( \displaystyle{ y = y_h + y_p = A + Be^{2t} + Ce^{-2t} + \frac{te^{2t}}{2} }\)

close solution

\( y^{(3)}-4y' = 5\cos(t) \)

Problem Statement

Use the method of undetermined coefficients to find the general solution to \( y^{(3)}-4y' = 5\cos(t) \)

Final Answer

\( y_p = -\sin(t) \)

Problem Statement

Use the method of undetermined coefficients to find the general solution to \( y^{(3)}-4y' = 5\cos(t) \)

Solution

We found the homogeneous solution in the previous problem. So here we need to find the particular solution. Looking at the right side of the equation, the form of our solution is \(y_p = A\cos(t)+B\sin(t)\).

\(y_p = A\cos(t)+B\sin(t)\)

\(y'_p=-A\sin(t)+B\cos(t)\)

\(y''_p=-A\cos(t)-B\sin(t)\)

\(y^{(3)}_p=A\cos(t)-B\cos(t)\)

Now plug in the derivatives above into the original differential equation to solve for A and B.
\(y^{(3)}-4y' = (A\cos(t)-B\cos(t))-4(-A\sin(t)+B\cos(t)) = \) \(5A\sin(t)-5B\cos(t) = 5\cos(t))\) \(\to A=0, B=-1\)

Final Answer

\( y_p = -\sin(t) \)

close solution

\( y'' - 5y' + 6y = 10e^{2x} \)

Problem Statement

\( y'' - 5y' + 6y = 10e^{2x} \)

Final Answer

\( y = Ae^{3x} + Be^{2x} - 10xe^{2x} \)

Problem Statement

\( y'' - 5y' + 6y = 10e^{2x} \)

Solution

615 solution video

video by Dr Chris Tisdell

Final Answer

\( y = Ae^{3x} + Be^{2x} - 10xe^{2x} \)

close solution

\( y'' - 3y' - 4y = 2\sin(x) \)

Problem Statement

\( y'' - 3y' - 4y = 2\sin(x) \)

Final Answer

\(\displaystyle{ y = c_1e^{4x} + c_2e^{-x} - \frac{5}{17}\sin(x) + \frac{3}{17}\cos(x) }\)

Problem Statement

\( y'' - 3y' - 4y = 2\sin(x) \)

Solution

617 solution video

video by Khan Academy

Final Answer

\(\displaystyle{ y = c_1e^{4x} + c_2e^{-x} - \frac{5}{17}\sin(x) + \frac{3}{17}\cos(x) }\)

close solution

\( y'' + y' - 2y = 3e^{-2t} \)

Problem Statement

Calculate \( y(t) = y_p + y_h \) of \( y'' + y' - 2y = 3e^{-2t} \) using undetermined coefficients.

Solution

2208 solution video

close solution

\( y'' - 3y' - 4y = 4x^2 \)

Problem Statement

\( y'' - 3y' - 4y = 4x^2 \)

Final Answer

\(\displaystyle{ y = c_1e^{4x} + c_2e^{-x} - x^2 + \frac{3}{2}x - \frac{13}{8} }\)

Problem Statement

\( y'' - 3y' - 4y = 4x^2 \)

Solution

618 solution video

video by Khan Academy

Final Answer

\(\displaystyle{ y = c_1e^{4x} + c_2e^{-x} - x^2 + \frac{3}{2}x - \frac{13}{8} }\)

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\( y'' + 2y' = 24x + e^{-2x} \)

Problem Statement

\( y'' + 2y' = 24x + e^{-2x} \)

Final Answer

\(\displaystyle{ y = c_1 + c_2e^{-2x} + 6x^2 - 6x - \frac{1}{2}xe^{-2x} }\)

Problem Statement

\( y'' + 2y' = 24x + e^{-2x} \)

Solution

622 solution video

video by Krista King Math

Final Answer

\(\displaystyle{ y = c_1 + c_2e^{-2x} + 6x^2 - 6x - \frac{1}{2}xe^{-2x} }\)

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\( y'' + 4y' + 4y = 130\cos(3x) \)

Problem Statement

\( y'' + 4y' + 4y = 130\cos(3x) \)

Final Answer

\(\displaystyle{ y(x) = e^{-2x}(c_1+c_2x) - \frac{50}{13}\cos(3x) + \frac{120}{13}\sin(3x) }\)

Problem Statement

\( y'' + 4y' + 4y = 130\cos(3x) \)

Solution

623 solution video

video by Krista King Math

Final Answer

\(\displaystyle{ y(x) = e^{-2x}(c_1+c_2x) - \frac{50}{13}\cos(3x) + \frac{120}{13}\sin(3x) }\)

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\( y'' + 7y' + 12y = 3e^{-3t} \)

Problem Statement

\( y'' + 7y' + 12y = 3e^{-3t} \)

Final Answer

\( y(t) = c_1e^{-4t} + c_2e^{-3t} + 3te^{-3t} \)

Problem Statement

\( y'' + 7y' + 12y = 3e^{-3t} \)

Solution

2080 solution video

video by MIP4U

Final Answer

\( y(t) = c_1e^{-4t} + c_2e^{-3t} + 3te^{-3t} \)

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\( 2y'' - y' - y = t^2+2t+5 \)

Problem Statement

\( 2y'' - y' - y = t^2+2t+5 \)

Solution

2212 solution video

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Advanced Problems

Calculate \(y_p\) of \(\displaystyle{ \frac{1}{2}y'' - y' + y = t\cos(t) }\).

Problem Statement

Calculate \(y_p\) of \(\displaystyle{ \frac{1}{2}y'' - y' + y = t\cos(t) }\).

Solution

2204 solution video

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