Ordinary Differential Equations: Problems with Solutions

Solution:
This is an equation with separable variables. For [tex]y\ne 0[/tex] we divide by [tex]y^{2}[/tex] and separate the variables:
[tex]\frac{dy}{y^{2}}=\frac{dx}{1+x^{2}}[/tex]
Integrating both sides:
[tex]-\frac{1}{y}=\arctan x+C\ \Rightarrow\ y=-\frac{1}{\arctan x+C}[/tex]
The constant of integration must stay in the denominator (writing [tex]y=-\frac{1}{\arctan x}+C[/tex] is a typical mistake). Check:
[tex]y'=\frac{1}{\left(\arctan x+C\right)^{2}}\,\frac{1}{1+x^{2}}=\frac{y^{2}}{1+x^{2}}[/tex]
The function [tex]y=0[/tex], lost when dividing by [tex]y^{2}[/tex], is also a solution of the equation.

Solution:
For [tex]x\ne 0[/tex] we divide by [tex]x[/tex]. The right-hand side depends only on [tex]\frac{y}{x}[/tex], so the equation is homogeneous:
[tex]y'=\frac{y}{x}+e^{\frac{y}{x}}[/tex]
Substitute [tex]y=xz[/tex], [tex]y'=z+xz'[/tex] and separate the variables:
[tex]z+xz'=z+e^{z}\ \Rightarrow\ xz'=e^{z}\ \Rightarrow\ e^{-z}\,dz=\frac{dx}{x}[/tex]
Integrating (we write the constant as [tex]-C[/tex] for convenience):
[tex]-e^{-z}=\ln|x|-C\ \Rightarrow\ e^{-z}=C-\ln|x|\ \Rightarrow\ z=-\ln\left(C-\ln|x|\right)[/tex]
where [tex]C-\ln|x|\gt 0[/tex]. Since [tex]y=xz[/tex],
[tex]y=-x\ln\left(C-\ln|x|\right)[/tex]
Check: here [tex]e^{\frac{y}{x}}=\frac{1}{C-\ln|x|}[/tex] and
[tex]y'=-\ln\left(C-\ln|x|\right)+\frac{1}{C-\ln|x|}\ \Rightarrow\ xy'=y+\frac{x}{C-\ln|x|}=y+xe^{\frac{y}{x}}[/tex]

Solution:
This is a linear first-order equation:
[tex]y'-y\tan x=\frac{1}{\cos x}[/tex]
An integrating factor is [tex]\mu=e^{-\int\tan x\,dx}=e^{\ln|\cos x|}=|\cos x|[/tex]. On every interval where [tex]\cos x\ne 0[/tex] the sign of [tex]\cos x[/tex] is constant, so we may take [tex]\mu=\cos x[/tex]. Multiplying by [tex]\cos x[/tex]:
[tex]y'\cos x-y\sin x=1\ \Rightarrow\ \left(y\cos x\right)'=1\ \Rightarrow\ y\cos x=x+C[/tex]
Hence
[tex]y=\frac{x+C}{\cos x}[/tex]
Check:
[tex]y'=\frac{\cos x+(x+C)\sin x}{\cos^{2}x}=\frac{1}{\cos x}+y\tan x[/tex]

Solution:
Since [tex]x\ne 0[/tex], we divide by [tex]x^{2}[/tex]. The left side becomes the derivative of the quotient [tex]\frac{y}{x}[/tex] (equivalently, [tex]\frac{1}{x}[/tex] is an integrating factor of the linear equation [tex]y'-\frac{y}{x}=x^{2}\cos x[/tex]):
[tex]\frac{xy'-y}{x^{2}}=\left(\frac{y}{x}\right)'=x\cos x[/tex]
Integrating by parts:
[tex]\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x+C[/tex]
Hence
[tex]\frac{y}{x}=x\sin x+\cos x+C\ \Rightarrow\ y=x^{2}\sin x+x\cos x+Cx[/tex]

Solution:
Put [tex]u=e^{y}\gt 0[/tex]. Then [tex]u'=e^{y}y'[/tex] and the equation becomes the linear equation
[tex]u'+u=1[/tex]
Multiplying by the integrating factor [tex]e^{x}[/tex]:
[tex]\left(ue^{x}\right)'=e^{x}\ \Rightarrow\ ue^{x}=e^{x}+C\ \Rightarrow\ u=1+Ce^{-x}[/tex]
Therefore, on an interval where [tex]1+Ce^{-x}\gt 0[/tex],
[tex]y=\ln\left(1+Ce^{-x}\right)[/tex]
Check:
[tex]y'=\frac{-Ce^{-x}}{1+Ce^{-x}}\ \Rightarrow\ y'+1=\frac{1}{1+Ce^{-x}}=e^{-y}\ \Rightarrow\ e^{y}(y'+1)=1[/tex]

Solution:
Write the equation in the form [tex]P\,dx+Q\,dy=0[/tex]:
[tex]\left(3x^{2}y^{2}+x\right)dx+2x^{3}y\,dy=0,\qquad P=3x^{2}y^{2}+x,\quad Q=2x^{3}y[/tex]
Since
[tex]\frac{\partial P}{\partial y}=6x^{2}y=\frac{\partial Q}{\partial x}[/tex]
the equation is exact, i.e. its left side is the total differential of a function [tex]U(x,y)[/tex]. From [tex]U_y=Q=2x^{3}y[/tex] we get [tex]U=x^{3}y^{2}+\varphi(x)[/tex]. Then [tex]U_x=P[/tex] gives
[tex]3x^{2}y^{2}+\varphi'(x)=3x^{2}y^{2}+x\ \Rightarrow\ \varphi'(x)=x\ \Rightarrow\ \varphi(x)=\frac{x^{2}}{2}[/tex]
The general integral is [tex]U(x,y)=C[/tex], i.e.
[tex]x^{3}y^{2}+\frac{x^{2}}{2}=C[/tex]

Solution:
Substituting [tex]t=0[/tex] into the equation:
[tex]f''(0)=4f'(0)-3f(0)+1=8-3+1=6[/tex]
The identity holds for every [tex]t[/tex] and its right-hand side is differentiable, so we may differentiate it:
[tex]f'''(t)=4f''(t)-3f'(t)\ \Rightarrow\ f'''(0)=4\times 6-3\times 2=18[/tex]
Differentiating once more:
[tex]f^{(4)}(t)=4f'''(t)-3f''(t)\ \Rightarrow\ f^{(4)}(0)=4\times 18-3\times 6=54[/tex]
Check by solving the equation: the characteristic equation [tex]r^{2}-4r+3=0[/tex] has roots [tex]1[/tex] and [tex]3[/tex], a particular solution is the constant [tex]\frac{1}{3}[/tex], and the initial conditions give
[tex]f(t)=\frac{2}{3}e^{3t}+\frac{1}{3},\qquad f^{(4)}(0)=\frac{2}{3}\times 81=54[/tex]

Solution:
This is a linear homogeneous equation with constant coefficients. Its characteristic equation and its roots are
[tex]\lambda^{3}+\lambda^{2}+\lambda=\lambda\left(\lambda^{2}+\lambda+1\right)=0\ \Rightarrow\ \lambda_1=0,\quad \lambda_{2,3}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i[/tex]
The root [tex]\lambda_1=0[/tex] gives the solution [tex]1[/tex], and the pair of complex conjugate roots [tex]\alpha\pm\beta i[/tex] with [tex]\alpha=-\frac{1}{2}[/tex], [tex]\beta=\frac{\sqrt{3}}{2}[/tex] gives [tex]e^{-\frac{x}{2}}\cos\frac{\sqrt{3}}{2}x[/tex] and [tex]e^{-\frac{x}{2}}\sin \frac{\sqrt{3}}{2}x[/tex]. These three functions form a fundamental system of solutions, so
[tex]y=C_1+e^{-\frac{x}{2}}\left(C_2\cos\frac{\sqrt{3}}{2}x+C_3\sin \frac{\sqrt{3}}{2}x\right)[/tex]
The equation is of third order, so its general solution must contain three arbitrary constants; forgetting the constant [tex]C_1[/tex] (coming from [tex]\lambda_1=0[/tex]) is a typical mistake.

Solution:
The characteristic equation of the homogeneous equation and its roots are
[tex]\lambda^{2}-\lambda+1=0\ \Rightarrow\ \lambda_{1,2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i[/tex]
so the general solution of the homogeneous equation is
[tex]y_0=e^{\frac{x}{2}}\left(C_1\cos\frac{\sqrt{3}}{2}x+C_2\sin \frac{\sqrt{3}}{2}x\right)[/tex]
The right-hand side is a polynomial of degree [tex]3[/tex] and [tex]0[/tex] is not a root of the characteristic equation, so we look for a particular solution of the form [tex]Y=ax^{3}+bx^{2}+cx+d[/tex]. Then
[tex]Y''-Y'+Y=ax^{3}+(b-3a)x^{2}+(6a-2b+c)x+(2b-c+d)[/tex]
Comparing the coefficients with those of [tex]x^{3}+6[/tex]:
[tex]a=1,\quad b=3a=3,\quad c=2b-6a=0,\quad d=6-2b+c=0[/tex]
Hence [tex]Y=x^{3}+3x^{2}[/tex], and the general solution is [tex]y=y_0+Y[/tex]:
[tex]y=e^{\frac{x}{2}}\left(C_1\cos\frac{\sqrt{3}}{2}x+C_2\sin \frac{\sqrt{3}}{2}x\right)+x^{3}+3x^{2}[/tex]

Solution:
The function is defined for every real [tex]x[/tex], because [tex]\sqrt{1+x^2}\gt |x|\ge -x[/tex], i.e. [tex]x+\sqrt{1+x^2}\gt 0[/tex]. First we differentiate the logarithm:
[tex]\left(\ln\left(x+\sqrt{1+x^2}\right)\right)'=\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}\left(x+\sqrt{1+x^2}\right)}=\frac{1}{\sqrt{1+x^2}}[/tex]
By the product rule
[tex]y'=\ln\left(x+\sqrt{1+x^2}\right)+\frac{x}{\sqrt{1+x^2}}-\frac{x}{\sqrt{1+x^2}}=\ln\left(x+\sqrt{1+x^2}\right)[/tex]
Hence
[tex]y''=\frac{1}{\sqrt{1+x^2}}=\left(1+x^2\right)^{-\frac12},\qquad y'''=-\frac12\left(1+x^2\right)^{-\frac32}\times 2x=-\frac{x}{\left(1+x^2\right)\sqrt{1+x^2}}[/tex]
Substituting into the left-hand side of the equation, for every [tex]x\in\mathbb{R}[/tex] we get
[tex](1+x^2)y'''+xy''=-\frac{x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}=0[/tex]
Therefore the function is a solution of the equation on [tex]\mathbb{R}[/tex].

Solution:
The characteristic equation of the homogeneous equation, its roots and the general solution [tex]Y[/tex] of the homogeneous equation:
[tex]\lambda^2+2\lambda-3=0,\quad \lambda_1=1,\ \lambda_2=-3,\qquad Y=C_1e^{x}+C_2e^{-3x}[/tex]
The right-hand side is [tex]8e^{x}[/tex] and [tex]1[/tex] is a simple root of the characteristic equation (resonance), so we look for a particular solution of the form [tex]\eta=Axe^{x}[/tex]:
[tex]\eta'=A(x+1)e^{x},\quad \eta''=A(x+2)e^{x},\quad \eta''+2\eta'-3\eta=Ae^{x}\left[(x+2)+2(x+1)-3x\right]=4Ae^{x}[/tex]
Hence [tex]4A=8[/tex], [tex]A=2[/tex], and the general solution and its derivative are
[tex]y=C_1e^{x}+C_2e^{-3x}+2xe^{x},\qquad y'=C_1e^{x}-3C_2e^{-3x}+2(x+1)e^{x}[/tex]
The initial conditions give
[tex]y(0)=C_1+C_2=0,\quad y'(0)=C_1-3C_2+2=6\ \Rightarrow\ C_1=1,\ C_2=-1[/tex]
Therefore [tex]y=e^{x}-e^{-3x}+2xe^{x}[/tex].

Solution:
The matrix of the system, its characteristic equation and its eigenvalues:
[tex]A=\begin{pmatrix}2&-3\\-1&0\end{pmatrix},\qquad \det(A-\lambda I)=(2-\lambda)(-\lambda)-3=\lambda^2-2\lambda-3=0,\quad \lambda_1=3,\ \lambda_2=-1[/tex]
For [tex]\lambda_1=3[/tex] the equations for an eigenvector [tex](a , b)[/tex] reduce to [tex]-a-3b=0[/tex], so we can take [tex](3 , -1)[/tex]; for [tex]\lambda_2=-1[/tex] they reduce to [tex]3a-3b=0[/tex], so we can take [tex](1 , 1)[/tex]. The general solution is
[tex]x=3C_1e^{3t}+C_2e^{-t},\qquad y=-C_1e^{3t}+C_2e^{-t}[/tex]
The initial conditions give
[tex]3C_1+C_2=1,\quad -C_1+C_2=1\ \Rightarrow\ C_1=0,\ C_2=1[/tex]
Hence [tex]x(t)=e^{-t}[/tex] and [tex]y(t)=e^{-t}[/tex]. Check:
[tex]x'=-e^{-t}=2e^{-t}-3e^{-t}=2x-3y,\qquad y'=-e^{-t}=-x[/tex]