1.5. Chapter 1 Exercises#
Exercise 1.1
Determine the type, order, degree, linearity and homogeneity of the following differential equations:
equation |
type |
order |
degree |
linearity |
homogeneity |
|---|---|---|---|---|---|
\(\displaystyle \diff{y}{x}=4x^2 - y\) |
_ |
_ |
_ |
_ |
_ |
\(\displaystyle \diff[2]{y}{x} - \left(\diff{y}{x}\right)^2 + 12xy=0\) |
_ |
_ |
_ |
_ |
_ |
\(\displaystyle \left(\diff{y}{x}\right)^2 + x \diff{y}{x} - 3y^2=0\) |
_ |
_ |
_ |
_ |
_ |
\(\displaystyle x\diff[2]{y}{x}-5\diff{y}{x}+3xy=\sin x\) |
_ |
_ |
_ |
_ |
_ |
\(\displaystyle \pdiff[2]{u}{x}+\pdiff[2]{u}{y}=0\) |
_ |
_ |
_ |
_ |
_ |
\(\displaystyle \pdiff{u}{t}+\pdiff{u}{x}=0\) |
_ |
_ |
_ |
_ |
_ |
\(\displaystyle \pdiff{u}{t}+u\pdiff{u}{x}=0\) |
_ |
_ |
_ |
_ |
_ |
\(\displaystyle \pdiff{u}{t}=\pdiff[2]{u}{x}\) |
_ |
_ |
_ |
_ |
_ |
Solution to
equation |
type |
order |
degree |
linearity |
homogeneity |
|---|---|---|---|---|---|
\(\displaystyle \diff{y}{x}=4x^2 - y\) |
ODE |
first |
first |
linear |
non-homogeneous |
\(\displaystyle \diff[2]{y}{x} - \left(\diff{y}{x}\right)^2 + 12xy=0\) |
ODE |
second |
first |
nonlinear |
homogeneous |
\(\displaystyle \left(\diff{y}{x}\right)^2 + x \diff{y}{x} - 3y^2=0\) |
ODE |
first |
second |
nonlinear |
homogeneous |
\(\displaystyle x\diff[2]{y}{x}-5\diff{y}{x}+3xy=\sin x\) |
ODE |
second |
first |
linear |
non-homogeneous |
\(\displaystyle \pdiff[2]{u}{x}+\pdiff[2]{u}{y}=0\) |
PDE |
second |
first |
linear |
homogeneous |
\(\displaystyle \pdiff{u}{t}+\pdiff{u}{x}=0\) |
PDE |
first |
first |
linear |
homogeneous |
\(\displaystyle \pdiff{u}{t}+u\pdiff{u}{x}=0\) |
PDE |
first |
first |
nonlinear |
homogeneous |
\(\displaystyle \pdiff{u}{t}=\pdiff[2]{u}{x}\) |
PDE |
second |
first |
linear |
homogeneous |
Exercise 1.2
Find the general solution to the following separable ODEs
\(\displaystyle y\dy = x \dx\)
\(\displaystyle \diff{y}{x}=y \ln y\)
\(\displaystyle y' = \frac{x^2}{y}\)
\(\displaystyle y'=\frac{x^2}{y(1+x^3)}\)
\(\displaystyle xy'=\sqrt{1-y^2}\)
\(\displaystyle \diff{y}{x}=\frac{x-e^{-x}}{y+e^{y}}\)
\(\displaystyle \diff{y}{x}=\frac{x^2}{1+y^2}\)
\(\displaystyle y'=\frac{2-e^x}{3+2y}\)
Solution to
\(\displaystyle y\dy = x \dx\)
\[\d \left(\frac{1}{2}y^2\right) = \d \left(\frac{1}{2}x^2\right) \]\[y^2=x^2+c \]\(\displaystyle \diff{y}{x}=y \ln y\)
\[ \frac{\dy}{y\ln y}=\dx \]\[ \frac{\d \ln y}{\ln y} = \dx\]Let \(u=\ln y\)
\[\frac{\d u}{u} = dx\]\[\d \ln u = \dx \]\[\ln u = x + c_1 \]\[ u = e^{x+c_1} = c e^x \]\[ \ln y = c e^x \]\(\displaystyle y' = \frac{x^2}{y}\)
\[ \diff{y}{x} = \frac{x^2}{y} \]\[ y \dy = x^2 \dx \]\[\d \left(\frac{1}{2}y^2\right)=\d \left(\frac{1}{3} x^3 \right)\]\[\frac{1}{2}y^2 = \frac{1}{3}x^3 +c \]\(\displaystyle y'=\frac{x^2}{y(1+x^3)}\)
\[ y \dy = \frac{x^2}{1+x^3} \dx \]\[ \d \left(\frac{1}{2}y^2\right) = \d \left(\frac{1}{3} \ln (1+x^3) \right)\]\[ \frac{1}{2}y^2 = \frac{1}{3} \ln (1+x^3) + c\]\(\displaystyle xy'=\sqrt{1-y^2}\)
\[ \frac{\dy}{\sqrt{1-y^2}}=\frac{\dx}{x} \]\[ \d (\sin^{-1} y) = \d (\ln x)\]\[ \sin^{-1} y = \ln x +c \]\[ y=\sin(\ln x + c ) \]\(\displaystyle \diff{y}{x}=\frac{x-e^{-x}}{y+e^{y}}\)
\[ (y+e^y) \dy = (x-e^{-x}) \dx \]\[ \d \left(\frac{1}{2}y^2 + e^y\right) = \d \left( \frac{1}{2}x^2 + e^{-x} \right)\]\[ \frac{1}{2}y^2 + e^y = \frac{1}{2}x^2 + e^{-x} + c \]\(\displaystyle \diff{y}{x}=\frac{x^2}{1+y^2}\)
\[ (1+y^2) \dy = x^2 \dx \]\[ \d \left(y+\frac{1}{3}y^3\right) = \d \left(\frac{1}{3}x^3\right)\]\[ y+\frac{1}{3}y^3 = \frac{1}{3}x^3 +c\]\(\displaystyle y'=\frac{2-e^x}{3+2y}\)
\[ (3+2y) \dy = (2-e^x) dx \]\[ \d (3y+y^2) = \d (2x-e^x)\]\[ 3y+y^2= 2x-e^x + c\]
Exercise 1.3
Solve the following initial value problems defined by separable ODEs
\(\displaystyle \diff{y}{x}=2xy\), \(y(0)=1\)
\(y^2 \dx + (x+1) \dy=0 \), \(y(0)=1\)
\(\displaystyle \diff{P}{t}=rP\left(1-\frac{P}{M}\right)\), \(P(0)=P_0\), \(r\) and \(M\) are constants, and \(0<P\ll M\).
Solution to
\(\displaystyle \diff{y}{x}=2xy\), \(y(0)=1\)
\[\frac{\dy}{y}=2x \dx\]\[ \ln |y| = x^2 + c_1 \]\[ y=e^{x^2+c_1}=c e^{x^2}\]\[ y(0)=c=1\]\[ y=e^{x^2}\]\(y^2 \dx + (x+1) \dy=0 \), \(y(0)=1\)
\[ y^2 \dx = -(x+1) \dy \]\[ \frac{\dx}{x+1}=-y^{-2} \dy\]\[ \d \ln (x+1) = \d y^{-1}\]\[ y^{-1} = \ln (x+1) + c\]\[ y= \frac{1}{\ln (x+1)+c}\]\[ y(0)=\frac{1}{\ln 1 + c} = 1\]\[ c=1\]\[ y= \frac{1}{\ln (x+1)+1} \]\(\displaystyle \diff{P}{t}=rP\left(1-\frac{P}{M}\right)\), \(P(0)=P_0\), \(r\) and \(M\) are constants, and \(0<P\ll M\).
\[ \frac{1}{P \left(1-\frac{P}{M}\right)} \d P = r \d t \]\[ \left(\frac{1}{P} + \frac{1}{M-P}\right) \d P = r \d t \]\[ \d \ln P - d \ln (M-P) = \d (rt)\]\[ \d \ln \left(\frac{P}{M-P}\right) = \d (rt)\]\[ \ln \left(\frac{P}{M-P}\right) = rt + c_1 \]\[ \frac{P}{M-P} = c e^{rt} \]\[ P (1+c e^{rt}) = c e^{rt} M\]\[ P = \frac{c e^{rt} M}{1+c e^{rt}}\]\[ P(0) = \frac{cM}{1+c}=P_0\]\[ c(M-P_0)=P_0\]\[ c=\frac{P_0}{M-P_0}\]\[ P = \frac{\frac{P_0}{M-P_0} e^{rt} M}{1+\frac{P_0}{M-P_0} e^{rt}}\]\[ P = \frac{M P_0 e^{rt}}{M+P_0 (e^{rt}-1)}\]
Exercise 1.4
Solve the following homogeneous ODEs for \(y=y(x)\), \(x\in\R\)
\(y''+y=0 \)
\(y'''+3y''+3y'+y=0\)
\(2y''-5y'-3y=0\)
\(y''-10y'+25y=0\)
\(20y''+4y'+y=0\)
Solution to
\(y''+y=0 \)
\[z^2+1=0 \]\[z_{1, 2} = 0\pm \i\]\[\begin{split} y & = e^{0x} (c_1 \cos x + c_2 \sin x) \\ & = c_1 \cos x + c_2 \sin x \end{split}\]\(y'''+3y''+3y'+y=0\)
\[ z^3+3z^2+3z+1 = 0 \]\[\begin{split} & (z^3+1) + (3z^2+3z)\\ = & (z+1)(z^2-z+1) + 3z(z+1) \\ = & (z+1)(z^2+2z+1) \\ = & (z+1)^3 \\ = & 0 \end{split}\]\[z_{1, 2, 3} = -1 \]\[ y=c_1 e^{-x}+c_2 x e^{-x} + c_3 x^2 e^{-x}\]\(2y''-5y'-3y=0\)
\[2z^2-5z-3 = 0\]\[(2z+1)(z-3)=0 \]\[z_{1}=-\tfrac{1}{2}, \quad z_2=3\]\[ y=c_1 e^{-\frac{1}{2}x} + c_2 e^{3x}\]\(y''-10y'+25y=0\)
\[z^2-10z+25=0\]\[(z-5)^2=0\]\[z_{1,2}=5\]\[y=c_1 e^{5x} + c_2 x e^{5x}\]\(20y''+4y'+y=0\)
\[20z^2+4z+1=0\]\[\begin{split} z_{1,2} & =\frac{-4\pm \sqrt{4^2-4(20)(1)}}{2(20)} \\ & = \frac{-4\pm 8 \i}{40} \\ & = -0.1 \pm 0.2 \i \end{split}\]\[ y=e^{-0.1x} (c_1 \cos 0.2x + c_2 \sin 0.2x) \]
Exercise 1.5
Solve the following non-homogeneous ODEs for \(y=y(x)\), \(x \in \R\)
\(y''+3y'+2y=2e^{3x}\)
\(y''+3y'+2y=x^2\)
\(y'''+6y''+11y'+6y=\cos 5x\)
Solution to
\(y''+3y'+2y=2e^{3x}\)
\[y_h''+3y_h'+2y_h=0\]\[z^2+3z+2=0\]\[z_1=-1, \quad z_2 = -2\]\[y_h=c_1 e^{-x} + c_2 e^{-2x}\]\[y_p=Ae^{3x}\]\[y_p'=3Ae^{3x}\]\[y_p''=9Ae^{3x}\]\[\begin{split} & y_p''+3y_p'+2y_p \\ = & ~ 9Ae^{3x} + 3 (3Ae^{3x}) + 2 (Ae^{3x}) \\ = & ~ 20 Ae^{3x} \\ = & ~ 2 e^{3x} \end{split}\]\[A=\tfrac{1}{10}\]\[\begin{split} y & = y_h + y_p \\ & =c_1 e^{-x} + c_2 e^{-2x} + \tfrac{1}{10}e^{3x} \end{split}\]\(y''+3y'+2y=x^2\)
\[y_h''+3y_h'+2y_h=0\]\[z^2+3z+2=0\]\[z_1=-1, \quad z_2 = -2\]\[y_h=c_1 e^{-x} + c_2 e^{-2x}\]\[y_p=Ax^2+Bx+C\]\[y_p'=2Ax+B\]\[y_p''=2A\]\[\begin{split} & y_p'' + 3y_p'+2y_p \\ = & ~ 2A+3(2Ax+B)+2(Ax^2+Bx+C) \\ = & ~ 2Ax^2 + (6A+2B)x + (2A+3B+2C) \\ = & x^2 \end{split}\]\[A=\tfrac{1}{2}, \quad B=-\tfrac{3}{2}, \quad C=\tfrac{7}{4}\]\[y_p=\tfrac{1}{2}x^2-\tfrac{3}{2}x+\tfrac{7}{4}\]\[\begin{split} y & = y_h + y_p \\ & =c_1 e^{-x} + c_2 e^{-2x} + \tfrac{1}{2}x^2-\tfrac{3}{2}x+\tfrac{7}{4} \end{split}\]\(y'''+6y''+11y'+6y=\cos 5x\)
\[z^3+6z^2+11z+6=0\]\[\begin{split} & z^3+6z^2+11z+6 \\ = & ~ z^3+6z^2+5z+6z+6 \\ = & ~ z(z^2+6z+5)+6(z+1) \\ = & ~ z(z+1)(z+5)+6(z+1) \\ = & ~ (z+1)[z(z+5)+6] \\ = & ~ (z+1)(z^2+5z+6) \\ = & ~ (z+1)(z+2)(z+3) \\ = & ~ 0 \end{split}\]\[ z_1 = -1, \quad z_2 = -2, \quad z_3=-3 \]\[y_h=c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{-3x}\]\[y_p = A \cos 5x + B \sin 5x\]\[y_p'= -5A \sin 5x + 5B \cos 5x\]\[y_p''= -25A \cos 5x - 25B \sin 5x\]\[y_p'''= 125A \sin 5x - 125B \cos 5x\]\[\begin{split} & y_p'''+6y_p''+11y_p'+6y_p \\ = &~(125A \sin 5x - 125B \cos 5x) \\ & + 6(-25A \cos 5x - 25B \sin 5x) \\ & + 11(-5A \sin 5x + 5B \cos 5x) \\ & + 6(A \cos 5x + B \sin 5x) \\ = & (70A-144B) \sin 5x + (-144A-70B) \cos 5x \\ = & \cos 5x \end{split}\]\[ A=-\tfrac{36}{6409}, \quad B=-\tfrac{35}{12818} \]\[y=c_1 e^{-x}+c_2 e^{-2x}+c_3 e^{-3x}-\tfrac{36}{6409}\cos 5x - \tfrac{35}{12818}\sin 5x\]
Exercise 1.6
Solve the following initial value problems for \(y=y(x)\), \(x \in [0, \infty)\)
\(y'=y+x\), \(y(0)=1\)
\(y'=-y+x+1\), \(y(0)=1\)
\(y'=-40y+40x+1\), \(y(0)=4\)
\(y'=-100y+\cos x\), \(y(0)=1\)
\(y''=-20y'-19y\), \(y(0)=2\), \(y'(0)=-20\)
Solution to
\(y'=y+x\), \(y(0)=1\)
\[y'+y=x\]\[z-1=0\]\[z=1\]\[y_h = c e^x\]\[y_p=Ax+B\]\[y_p'=A\]\[y_p'-y_p = A- (Ax+B) = x\]\[A=-1, \quad B=-1 \]\[y_p=-x-1\]\[y=y_h+y_p=ce^x-x-1\]\[y(0)=c-1=1 \]\[c=2\]\[y=y_h+y_p=2e^x-x-1\]\(y'=-y+x+1\), \(y(0)=1\)
\[y'+y=x+1\]\[z+1=0\]\[z=-1\]\[y_h=ce^{-x}\]\[y_p=Ax+B\]\[y_p'+y_p=A+(Ax+B)=x+1\]\[A=1, \quad B=0\]\[y_p=x\]\[y=y_h+y_p=ce^{-x}+x\]\[y(0)=c+0=1\]\[c=1\]\[y=e^{-x}+x\]\(y'=-40y+40x+1\), \(y(0)=4\)
\[y'+40y=40x+1\]\[z+40=0\]\[z=-40\]\[y_h=ce^{-40x}\]\[y_p=Ax+B\]\[y_p'+40y_p = A + 40(Ax+B)=40x+1\]\[A=1, \quad B=0\]\[y_p=x\]\[y=y_h+y_p=ce^{-40x}+x\]\[y(0)=c=4\]\[y=4e^{-40x}+x \]\(y'=-100y+\cos x\), \(y(0)=1\)
\[y'+100y=\cos x\]\[z+100=0\]\[z=-100\]\[y_h = c e^{-100x}\]\[y_p = A \cos x + B \sin x\]\[y_p'=-A \sin x + B \cos x\]\[\begin{split}y_p'+100 y_p & = (-A\sin x+ B \cos x)+100(A\cos x+B\sin x) \\ & = (100B-A) \sin x + (100A+B) \cos x \\ & = \cos x \end{split}\]\[\begin{split} & 100B-A = 0\\ & 100A+B = 1 \end{split}\]\[A=\tfrac{100}{10001}, \quad B=\tfrac{1}{10001} \]\[y_p = \tfrac{100}{10001} \cos x + \tfrac{1}{10001} \sin x\]\[y=y_h+y_p = c e^{-100x}+ \tfrac{100}{10001} \cos x + \tfrac{1}{10001} \sin x \]\[y(0)=c+\tfrac{100}{10001}=1\]\[c=\tfrac{9901}{10001}\]\[y=\tfrac{9901}{10001}e^{-100x} + \tfrac{100}{10001} \cos x + \tfrac{1}{10001} \sin x \]\(y''=-20y'-19y\), \(y(0)=2\), \(y'(0)=-20\)
\[y''+20y'+19y=0\]\[z^2+20z+19=0\]\[z_1 = -1, \quad z_2 = -19\]\[y=c_1 e^{-x}+c_2 e^{-19x}\]\[y'=-c_1 e^{-x}-19 c_2 e^{-19x}\]\[\begin{split} y(0) & = c_1 + c_2 = 2\\ y'(0)& = -c_1 -19 c_2 = -20 \end{split}\]\[c_1= 1, \quad c_2 = 1\]\[ y=e^{-x} + e^{-19x} \]