Ordinary Differential Equations: Difficult Problems with Solutions

Solution:
By Newton's second law (with the positive direction downwards):
[tex]m\frac{dv}{dt}=mg-kv^2,\qquad v(0)=0[/tex]
Let [tex]V=\sqrt{\frac{mg}{k}}[/tex]; then [tex]\frac{dv}{dt}=\frac{k}{m}\left(V^2-v^2\right)[/tex]. The constants [tex]v=\pm V[/tex] are solutions, so by uniqueness the solution with [tex]v(0)=0[/tex] satisfies [tex]|v|\lt V[/tex] for all [tex]t\ge 0[/tex]. Hence we may separate the variables and integrate, using [tex]v(0)=0[/tex]:
[tex]\frac{dv}{V^2-v^2}=\frac{k}{m}\,dt\ \Rightarrow\ \frac{1}{2V}\ln\frac{V+v}{V-v}=\frac{k}{m}\,t\ \Rightarrow\ \frac{V+v}{V-v}=e^{\frac{2kV}{m}t}[/tex]
Solving for [tex]v[/tex]:
[tex]v(t)=V\,\frac{e^{\frac{2kV}{m}t}-1}{e^{\frac{2kV}{m}t}+1}=V\,\frac{1-e^{-\frac{2kV}{m}t}}{1+e^{-\frac{2kV}{m}t}}[/tex]
Since [tex]e^{-\frac{2kV}{m}t}\to 0[/tex] as [tex]t\to\infty[/tex],
[tex]\lim_{t\to\infty}v(t)=V=\sqrt{\frac{mg}{k}}[/tex]
This is the terminal velocity, at which gravity and resistance balance: [tex]mg=kv^2[/tex].

Solution:
Rewrite the equation as
[tex](1-x^2)\,y'=xy-xy^2=xy(1-y)[/tex]
This is a Bernoulli equation, but its variables also separate. Near the point [tex]M[/tex] we have [tex]|x|\lt 1[/tex] and [tex]y(1-y)\ne 0[/tex], so
[tex]\frac{dy}{y(1-y)}=\frac{x\,dx}{1-x^2}[/tex]
Since [tex]\frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}[/tex], integration gives
[tex]\ln\left|\frac{y}{1-y}\right|=-\frac{1}{2}\ln\left(1-x^2\right)+c\ \Rightarrow\ \frac{y}{1-y}=\frac{C}{\sqrt{1-x^2}}[/tex]
The point [tex]M\left(0 , \frac{1}{2}\right)[/tex] gives [tex]C=1[/tex]. Then
[tex]y\sqrt{1-x^2}=1-y\ \Rightarrow\ y=\frac{1}{1+\sqrt{1-x^2}},\qquad -1\lt x\lt 1[/tex]

Solution:
The equation does not contain [tex]y[/tex], so we lower the order with [tex]y'=p(x)[/tex], [tex]y''=p'[/tex]:
[tex]p'=p^2+9[/tex]
We separate the variables ([tex]p^2+9\gt 0[/tex]):
[tex]\frac{dp}{p^2+9}=dx\ \Rightarrow\ \frac{1}{3}\arctan\frac{p}{3}=x+c\ \Rightarrow\ p=3\tan \left(3x+C_1\right),\quad C_1=3c[/tex]
Integrating once more:
[tex]y=\int 3\tan \left(3x+C_1\right)dx=-\ln\left|\cos\left(3x+C_1\right)\right|+C_2[/tex]
Check:
[tex]y'=3\tan \left(3x+C_1\right),\qquad y''=\frac{9}{\cos^2\left(3x+C_1\right)}=9+9\tan ^2\left(3x+C_1\right)=9+(y')^2[/tex]

Solution:
This is an Euler-type equation. Since [tex]2x+1\gt 0[/tex], we substitute [tex]2x+1=e^{t}[/tex], i.e. [tex]t=\ln(2x+1)[/tex] and [tex]\frac{dt}{dx}=\frac{2}{2x+1}=2e^{-t}[/tex]. Denoting the derivatives with respect to [tex]t[/tex] by dots, we get
[tex]y'=\dot y\,\frac{dt}{dx}=2e^{-t}\dot y,\qquad y''=\frac{d}{dt}\left(2e^{-t}\dot y\right)\times \frac{dt}{dx}=4e^{-2t}\left(\ddot y-\dot y\right)[/tex]
Since [tex](2x+1)^2=e^{2t}[/tex], substitution into the equation gives
[tex]4\left(\ddot y-\dot y\right)-4\dot y+4y=0\ \Leftrightarrow\ \ddot y-2\dot y+y=0[/tex]
Its characteristic equation [tex]r^2-2r+1=0[/tex] has the double root [tex]r=1[/tex], so [tex]y=(C_1+C_2t)e^{t}[/tex]. Returning to [tex]x[/tex]:
[tex]y=(2x+1)\left(C_1+C_2\ln(2x+1)\right)[/tex]

Solution:
The equation is not linear in [tex]y[/tex], but it becomes linear if [tex]x[/tex] is regarded as a function of [tex]y[/tex]. The equation implies [tex]y'\ne 0[/tex], so [tex]y'=\frac{1}{x'(y)}[/tex] and
[tex]\frac{dx}{dy}=x+2y\ \Leftrightarrow\ x'-x=2y[/tex]
We multiply by the integrating factor [tex]e^{-y}[/tex] and integrate by parts:
[tex]\left(xe^{-y}\right)'=2ye^{-y}\ \Rightarrow\ xe^{-y}=-2ye^{-y}-2e^{-y}+C\ \Rightarrow\ x=Ce^{y}-2y-2[/tex]
The initial condition means that [tex]x=0[/tex] when [tex]y=-1[/tex]:
[tex]0=Ce^{-1}+2-2\ \Rightarrow\ C=0,\qquad x=-2y-2\ \Leftrightarrow\ y=-\frac{x}{2}-1[/tex]
Check: [tex]x+2y=-2[/tex] and [tex]y'=-\frac{1}{2}[/tex], so [tex](x+2y)y'=1[/tex]; also [tex]y(0)=-1[/tex].

Solution:
Where [tex]x\ne 0[/tex] and [tex]2x+y\ne 0[/tex], the equation is homogeneous:
[tex]y'=\frac{x+2y}{2x+y}=\frac{1+2\frac{y}{x}}{2+\frac{y}{x}}[/tex]
Put [tex]y=ux[/tex], [tex]y'=u+xu'[/tex]:
[tex]u+xu'=\frac{1+2u}{2+u}\ \Rightarrow\ xu'=\frac{1-u^2}{2+u}[/tex]
For [tex]u\ne\pm1[/tex] we separate the variables and decompose into partial fractions:
[tex]\frac{2+u}{1-u^2}\,du=\frac{dx}{x},\qquad \frac{2+u}{(1-u)(1+u)}=\frac{3}{2(1-u)}+\frac{1}{2(1+u)}[/tex]
Integrating, multiplying by [tex]2[/tex] and exponentiating:
[tex]-\frac{3}{2}\ln|1-u|+\frac{1}{2}\ln|1+u|=\ln|x|+C_1\ \Rightarrow\ \frac{1+u}{(1-u)^3}=Cx^2[/tex]
Returning to [tex]u=\frac{y}{x}[/tex]:
[tex]\frac{(x+y)x^2}{(x-y)^3}=Cx^2\ \Rightarrow\ x+y=C(x-y)^3[/tex]
The excluded cases [tex]u=\pm1[/tex] give [tex]y=-x[/tex] (the family with [tex]C=0[/tex]) and [tex]y=x[/tex] (obtained formally for [tex]C\to\infty[/tex]); both satisfy the equation. The general integral is [tex]x+y=C(x-y)^3[/tex].

Solution:
This is a Bernoulli equation with exponent [tex]3[/tex]; [tex]y=0[/tex] is a solution. For [tex]y\ne 0[/tex] we divide by [tex]y^3[/tex] and substitute [tex]z=y^{-2}[/tex], [tex]z'=-2y^{-3}y'[/tex]:
[tex]2y^{-3}y'\sin x+y^{-2}\cos x=x\cos x-\sin x\ \Rightarrow\ -z'\sin x+z\cos x=x\cos x-\sin x[/tex]
This equation is linear in [tex]z[/tex]. On an interval where [tex]\sin x\ne 0[/tex] we divide by [tex]-\sin ^2x[/tex] and recognize both sides as derivatives of quotients:
[tex]\frac{z'\sin x-z\cos x}{\sin ^2x}=\frac{\sin x-x\cos x}{\sin ^2x}\ \Rightarrow\ \left(\frac{z}{\sin x}\right)'=\left(\frac{x}{\sin x}\right)'[/tex]
Hence
[tex]\frac{z}{\sin x}=\frac{x}{\sin x}+C\ \Rightarrow\ z=x+C\sin x\ \Rightarrow\ \frac{1}{y^2}=x+C\sin x[/tex]
The general integral is [tex]\frac{1}{y^2}=x+C\sin x[/tex] (together with the solution [tex]y=0[/tex]).

Solution:
Here [tex]P=xy+y^{2}[/tex], [tex]Q=xy+1[/tex], [tex]P_y=x+2y[/tex], [tex]Q_x=y[/tex], so the equation is not exact. An integrating factor [tex]\mu(y)[/tex] must satisfy [tex](\mu P)_y=(\mu Q)_x[/tex], i.e.
[tex]\frac{\mu'}{\mu}=\frac{Q_x-P_y}{P}=\frac{-(x+y)}{y(x+y)}=-\frac{1}{y}[/tex]
The right side depends only on [tex]y[/tex], so such a factor exists: [tex]\mu=\frac{1}{y}[/tex]. Multiplying by it gives an exact equation:
[tex](x+y)\,dx+\left(x+\frac{1}{y}\right)dy=0,\qquad \frac{\partial}{\partial y}(x+y)=1=\frac{\partial}{\partial x}\left(x+\frac{1}{y}\right)[/tex]
We look for [tex]U[/tex] with [tex]U_x=x+y[/tex], [tex]U_y=x+\frac{1}{y}[/tex]:
[tex]U=\frac{x^{2}}{2}+xy+\psi(y),\quad U_y=x+\psi'(y)=x+\frac{1}{y}\ \Rightarrow\ \psi(y)=\ln|y|[/tex]
The general integral is [tex]\frac{x^{2}}{2}+xy+\ln|y|=C[/tex], i.e. [tex]x^{2}+2xy+2\ln|y|=C[/tex]. The function [tex]y=0[/tex], lost when multiplying by [tex]\frac{1}{y}[/tex], also satisfies the equation.

Solution:
The equation does not contain [tex]x[/tex] explicitly, so we lower its order with [tex]y'=p(y)[/tex], [tex]y''=p\frac{dp}{dy}[/tex]:
[tex](y-1)p\frac{dp}{dy}=2p^{2}[/tex]
The case [tex]p\equiv 0[/tex] gives constants, which contradict [tex]y'(0)=-1[/tex]. So we divide by [tex]p[/tex] and separate the variables:
[tex]\frac{dp}{p}=\frac{2\,dy}{y-1}\ \Rightarrow\ p=C_1(y-1)^{2}[/tex]
At [tex]x=0[/tex] we have [tex]y=2[/tex], [tex]p=-1[/tex], so [tex]C_1=-1[/tex] and [tex]y'=-(y-1)^{2}[/tex]. Separating the variables again:
[tex]\frac{dy}{(y-1)^{2}}=-dx\ \Rightarrow\ -\frac{1}{y-1}=-x+C_2[/tex]
From [tex]y(0)=2[/tex] we get [tex]C_2=-1[/tex], hence [tex]\frac{1}{y-1}=x+1[/tex] and
[tex]y=1+\frac{1}{x+1}=\frac{x+2}{x+1},\quad x\gt -1[/tex]
Check:
[tex]y'=-\frac{1}{(x+1)^{2}},\quad y''=\frac{2}{(x+1)^{3}},\quad (y-1)y''=\frac{2}{(x+1)^{4}}=2(y')^{2},\quad y(0)=2,\quad y'(0)=-1[/tex]

Solution:
The characteristic equation is
[tex]\lambda^{3}-3\lambda-2=(\lambda-2)(\lambda+1)^{2}=0[/tex]
Its roots are [tex]2[/tex] (simple) and [tex]-1[/tex] (double), so the general solution of the homogeneous equation is
[tex]y_0=C_1e^{2x}+(C_2+C_3x)e^{-x}[/tex]
The right-hand side is [tex]9e^{2x}[/tex] and [tex]2[/tex] is a simple root (resonance), so we seek a particular solution of the form [tex]Y=Axe^{2x}[/tex]:
[tex]Y'=A(1+2x)e^{2x},\quad Y''=A(4+4x)e^{2x},\quad Y'''=A(12+8x)e^{2x}[/tex]
[tex]Y'''-3Y'-2Y=A(12+8x-3-6x-2x)e^{2x}=9Ae^{2x}=9e^{2x}\ \Rightarrow\ A=1[/tex]
Therefore
[tex]y=C_1e^{2x}+(C_2+C_3x)e^{-x}+xe^{2x}[/tex]

Solution:
The characteristic equation [tex]\lambda^2+1=0[/tex] has roots [tex]\pm i[/tex], so the homogeneous equation has the general solution
[tex]y_0=C_1\cos x+C_2\sin x[/tex]
The right-hand side is a first-degree polynomial times [tex]\cos x[/tex], and [tex]i[/tex] is a simple root of the characteristic equation (resonance), so we seek a particular solution
[tex]\eta=x\left[(Ax+B)\cos x+(Mx+N)\sin x\right]=P(x)\cos x+Q(x)\sin x,\quad P=Ax^2+Bx,\ Q=Mx^2+Nx[/tex]
For such a function
[tex]\eta''+\eta=(P''+2Q')\cos x+(Q''-2P')\sin x=(2A+2N+4Mx)\cos x+(2M-2B-4Ax)\sin x[/tex]
Comparing with [tex]4x\cos x[/tex]:
[tex]4M=4,\ 2A+2N=0,\ -4A=0,\ 2M-2B=0\ \Rightarrow\ A=0,\ B=1,\ M=1,\ N=0[/tex]
So [tex]\eta=x\cos x+x^2\sin x[/tex] and the general solution is
[tex]y=C_1\cos x+C_2\sin x+x\cos x+x^2\sin x[/tex]

Solution:
The right-hand side is not of special form, so we use variation of constants (Lagrange's method). The homogeneous equation has the general solution [tex]C_1\cos x+C_2\sin x[/tex]; we seek [tex]y=C_1(x)\cos x+C_2(x)\sin x[/tex] with
[tex]\begin{cases}C_1'\cos x+C_2'\sin x=0\\ -C_1'\sin x+C_2'\cos x=\frac{1}{\sin x}\end{cases}[/tex]
The determinant of this system is the Wronskian [tex]\cos^2x+\sin ^2x=1[/tex], and
[tex]C_1'=-1,\quad C_2'=\frac{\cos x}{\sin x}\ \Rightarrow\ C_1=-x+A,\quad C_2=\ln|\sin x|+B[/tex]
[tex]y=A\cos x+B\sin x-x\cos x+\sin x\ln|\sin x|[/tex]
The initial point [tex]\frac{\pi}{2}[/tex] lies in [tex](0, \pi)[/tex], where [tex]\sin x\gt 0[/tex] and the right-hand side is continuous, so the solution is considered on this interval. Then [tex]y\left(\frac{\pi}{2}\right)=B=5[/tex] and
[tex]y'=-A\sin x+B\cos x+x\sin x+\cos x\ln(\sin x)\ \Rightarrow\ y'\left(\frac{\pi}{2}\right)=-A+\frac{\pi}{2}=\frac{\pi}{2}\ \Rightarrow\ A=0[/tex]
Therefore
[tex]y=5\sin x-x\cos x+\sin x\ln(\sin x),\quad 0\lt x\lt\pi[/tex]

Solution:
From the second equation [tex]x=y-\dot{y}[/tex], hence [tex]\dot{x}=\dot{y}-\ddot{y}[/tex]. Substituting into the first equation:
[tex]\dot{y}-\ddot{y}=3(y-\dot{y})+2y\ \Rightarrow\ \ddot{y}-4\dot{y}+5y=0[/tex]
The characteristic equation [tex]r^2-4r+5=0[/tex] has roots [tex]r=2\pm i[/tex] (the eigenvalues of the matrix of the system), therefore
[tex]y=e^{2t}\left(C_1\cos t+C_2\sin t\right)[/tex]
Then
[tex]\dot{y}=e^{2t}\left[(2C_1+C_2)\cos t+(2C_2-C_1)\sin t\right][/tex]
[tex]x=y-\dot{y}=e^{2t}\left[-(C_1+C_2)\cos t+(C_1-C_2)\sin t\right][/tex]
Direct substitution confirms that both equations of the system are satisfied.

Solution:
Let [tex]M(x, y)[/tex] be a point of the curve; for the triangle to exist we need [tex]y\ne 0[/tex] and [tex]y'\ne 0[/tex]. The tangent at [tex]M[/tex] is [tex]Y-y=y'(X-x)[/tex]; it meets the x-axis at [tex]T\left(x-\frac{y}{y'}, 0\right)[/tex], and the perpendicular from [tex]M[/tex] to the x-axis ends at [tex]P(x, 0)[/tex]. The triangle [tex]MPT[/tex] is right-angled at [tex]P[/tex] and its legs are
[tex]|MP|=|y|,\qquad |PT|=\left|\frac{y}{y'}\right|[/tex]
(the segment [tex]PT[/tex] is the subtangent), so the condition of the problem is
[tex]\frac{1}{2}\,|y|\left|\frac{y}{y'}\right|=a^2\ \Rightarrow\ \frac{y^2}{2|y'|}=a^2\ \Rightarrow\ y'=\pm\frac{y^2}{2a^2}[/tex]
Separating the variables and integrating:
[tex]\frac{dy}{y^2}=\pm\frac{dx}{2a^2}\ \Rightarrow\ -\frac{1}{y}=\pm\frac{x+C}{2a^2}\ \Rightarrow\ (x+C)\,y=\mp 2a^2[/tex]
Both signs are possible, so the required curves are the hyperbolas [tex](x+C)\,y=\pm 2a^2[/tex] with asymptotes [tex]y=0[/tex] and [tex]x=-C[/tex]. Check: for [tex]y=\frac{2a^2}{x+C}[/tex] we have [tex]y'=-\frac{2a^2}{(x+C)^2}[/tex] and
[tex]\frac{y^2}{2|y'|}=\frac{4a^4}{(x+C)^2}\times \frac{(x+C)^2}{4a^2}=a^2[/tex]

Solution:
Put [tex]y'=p[/tex], so [tex]y=xp-4p^3[/tex]. Differentiate with respect to [tex]x[/tex] and use [tex]y'=p[/tex]:
[tex]p=p+xp'-12p^2p'\ \Rightarrow\ p'\left(x-12p^2\right)=0[/tex]
If [tex]p'=0[/tex], then [tex]p=C[/tex] and we get the general solution [tex]y=Cx-4C^3[/tex] (a family of straight lines). If [tex]x=12p^2[/tex], then [tex]y=12p^3-4p^3=8p^3[/tex]. Eliminate the parameter [tex]p[/tex]:
[tex]x^3=1728p^6,\quad y^2=64p^6\ \Rightarrow\ x^3=27y^2[/tex]
Check for the branch [tex]y=\frac{x\sqrt{x}}{3\sqrt{3}}[/tex], [tex]x\ge 0[/tex]: here [tex]y'=\frac{\sqrt{x}}{2\sqrt{3}}[/tex] and
[tex]xy'-4(y')^3=\frac{x\sqrt{x}}{2\sqrt{3}}-\frac{x\sqrt{x}}{6\sqrt{3}}=\frac{x\sqrt{x}}{3\sqrt{3}}=y[/tex]
(the branch with the minus sign is analogous). This curve is not a straight line, so it is not obtained from the general solution for any [tex]C[/tex]; it is the envelope of the lines [tex]y=Cx-4C^3[/tex] (each of them touches it at the point with abscissa [tex]x=12C^2[/tex]). Hence the singular solution is [tex]27y^2=x^3[/tex].

Solution:
Since [tex](yy')'=yy''+(y')^{2}[/tex], the equation can be written as [tex](yy')'=-1[/tex]. We integrate it twice, using [tex]yy'=\frac{1}{2}\left(y^{2}\right)'[/tex]:
[tex]yy'=-x+C_1\ \Rightarrow\ \left(y^{2}\right)'=-2x+2C_1\ \Rightarrow\ y^{2}=-x^{2}+2C_1x+C_2[/tex]
[tex](x-C_1)^{2}+y^{2}=C_1^{2}+C_2[/tex]
A solution never vanishes: at a point where [tex]y=0[/tex] the equation would give [tex](y')^{2}+1=0[/tex], which is impossible. Hence [tex]C_1^{2}+C_2=R^{2}\gt 0[/tex], and every integral curve lies on the circle with centre [tex](C_1, 0)[/tex] on the x-axis and radius [tex]R[/tex]; since [tex]y[/tex] keeps its sign, the curve is an arc of the upper or of the lower half of this circle.
Conversely, every such arc [tex]y=\pm\sqrt{R^{2}-(x-x_0)^{2}}[/tex], [tex]|x-x_0|\lt R[/tex], is a solution: differentiating [tex](x-x_0)^{2}+y^{2}=R^{2}[/tex] twice gives
[tex](x-x_0)+yy'=0\ \Rightarrow\ 1+(y')^{2}+yy''=0[/tex]
This proves the statement.