4  L14: Undetermined Coefficients I (5.4)

4.1 Recall

\[y''+py'+qy=F(x);\qquad p,q\in \mathbb{R}\]

has general solution:

\[y=c_1y_1+c_2y_2+y_p\] where \(\{y_1,y_2\}\) is a fundamental set of solutions to:

\[y''+py'+qy=0\]

and \(y_p\) is a particular solution to

\[y''+py'+qy=F(x)\]

4.2 Method of Undetermined Coefficients

\[y''+py'+qy=F(x) \tag{4.1}\]

STEP 1. Solve for the roots of \(r^2+pr+q=0\) and then state the solution to the complementary equation: \[y_c=c_1y_1+c_2y_2\]

STEP 2. Find roots associated with \(F(x)\) and combine them with the roots of the characteristic equation to find:

\[y_q=(c_1y_1+c_2y_2)+(c_3y_3+\cdots+c_ny_n)\]

STEP 3. Substitute \(y_p=y_q-y_c\) into Equation 5.1 to determine \(c_3,\cdots,c_n\).

STEP 4. General solution:

\[y=c_1y_1+c_2y_2+y_p\]

4.3 Example eeeeasy

\[y''-7y'+12y=4e^{2x}\]

STEP 1:

Complementary equation:

\[y''-7y'+12y=0\] Characteristic equation:

\[r^2-7r+12=0\] \[\Rightarrow (r-3)(r-4)=0\] \[\Rightarrow r=3,\quad r=4\] \[\Rightarrow y_c=c_1e^{3x}+c_2e^{4x}\]

STEP 2:

Roots for \(F(x)=4e^{2x},\qquad r=2\)

Then All roots: \(\boxed{r=3,4,2}\)1

\[\Rightarrow y_q=c_1e^{3x}+c_2e^{4x}+c_3e^{2x}\] STEP 3:

\[\Rightarrow y_p=y_q-y_c \rightarrow \boxed{y_p=c_3e^{2x}}\] \[\Rightarrow \text{we need to find }c_3: \] \[y_p''-7y_p'+12y_p=4e^{2x}\] \[4c_3e^{2x}-7c_3\cdot2e^{2x}+12\cdot c_3e^{2x}=4e^{2x}\] \[ (4-14+12)c_3=4\Rightarrow \boxed{c_3=2}\]

STEP 4:

Finally the General Solution for the 2OLNHCC is:

\[\boxed{\boxed{y=c_1e^{3x}+c_2e^{4x}+2e^{2x}}}\]

4.4 Example easy

\[y''-7y'+12y=5e^{4x}\]

STEP 1:

\[(r_1-4)(r_2-3)\Rightarrow y_c=c_1e^{3x}+c_2e^{4x}\] STEP 2:

\[\boxed{r_3=4}\,\text{Repeated!}\Rightarrow y_q=c_1e^{3x}+c_2e^{4x}+\boxed{c_3xe^{4x}}\] STEP 3:

\[y_p=y_q-y_c\Rightarrow \boxed{y_p=c_3xe^{4x}}\] \[\rightarrow y_p'=c_3e^{4x}+4c_3xe^{4x}=c_3e^{4x}(1+4x)\]

\[\rightarrow y_p''=4c_3e^{4x}+4c_3e^{4x}+16c_3xe^{4x}=8c_3e^{4x}(1+2x)\] \[\Rightarrow y_p''-7y_p'+12y_p=5e^{4x}\qquad\text{(becomes:)}\]

\[8c_3e^{4x}(1+2x)-7(c_3e^{4x}(1+4x))+12(c_3xe^{4x})=5e^{4x}\]

\[(8-7)c_3=5\rightarrow \boxed{c_3=5}\]

STEP 4:

\[\boxed{\boxed{y=c_1e^{3x}+c_2e^{4x}+5xe^{4x}}}\]

4.5 Example A-type

\[y''-8y'+16y=2e^{4x}\]

  1. \[y_c=c_1e^{4x}+c_2xe^{4x}\qquad\text{(repeated roots)}\]
  2. I cant add another \(xe^{4x}\), because that is linear dependent, then I assume \(u(x)e^{4x}\) is a soln, and see what are the conditions \(u(x)\) need to satisfy, i.e.

\[y''-8y'+16y=2e^{4x}\] \[(u(x)e^{4x})''-8(u(x)e^{4x})'+16(u(x)e^{4x})=2e^{4x}\]

\[(ue^{4x})''-8(u'e^{4x}+4ue^{4x})+16(ue^{4x})=2e^{4x}\] \[(u'e^{4x}+4ue^{4x})'-8(u'e^{4x}+4ue^{4x})+16(ue^{4x})=2e^{4x}\] \[(u''e^{4x}+4u'e^{4x}+4u'e^{4x}+16ue^{4x})-8(u'e^{4x}+4ue^{4x})+16(ue^{4x})=2e^{4x}\]

\[(u''+8u'+16u)-8(u'+4u)+16(u)=2\] \[u''=2\] \[\Rightarrow u'=2x \Rightarrow \boxed{u=x^2}\] 3. \[\Rightarrow y_q=c_1e^{4x}+c_2xe^{4x}+c_3x^2e^{4x}\]

\[\Rightarrow y_p=c_3x^2e^{4x}\] \[\Rightarrow y_p'=2c_3xe^{4x}+4c_3x^2e^{4x}\] \[\Rightarrow y_p''=2c_3e^{4x}+8c_3xe^{4x}+8c_3xe^{4x}+16c_3x^2e^{4x}\] \[\Rightarrow y_p''=2c_3e^{4x}+16c_3xe^{4x}+16c_3x^2e^{4x}\] \[(2c_3+16c_3x+16c_3x^2)-8(2c_3x+4c_3x^2)+16(c_3x^2)=2\] \[2c_3+(16c_3-16c_3)x+(16c_3-32c_3+16c_3)x^2=2\] \[\Rightarrow \boxed{c_3=1}\]

  1. Then the general equation is:

\[\boxed{\boxed{y=c_1e^{4x}+c_2xe^{4x}+x^2e^{4x}}}\]

4.6 More examples

Click here!

4.7 References

  1. https://quarto.org/docs/authoring/cross-references.html#theorems-and-proofs
  2. https://quarto.org/docs/presentations/revealjs/advanced.html
  3. https://quarto.org/docs/presentations/revealjs/presenting.html#chalkboard
  4. LibreTexts, 5.4: The Method of Undetermined Coefficients I, from William F. Trench.
  5. Undetermined coefficients 1 | Second order differential equations | Khan Academy

  1. all distinct, that is good!↩︎