# 2.4E: Transformation of Nonlinear Equations into Separable Equations (Exercises)

- Page ID
- 18250

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In Exercises [exer:2.4.1}– [exer:2.4.4} solve the given Bernoulli equation.

[exer:2.4.1] \(y'+y=y^2\)

[exer:2.4.2] \( {7xy'-2y=-{x^2 \over y^6}}\)

[exer:2.4.3] \(x^2y'+2y=2e^{1/x}y^{1/2}\)

[exer:2.4.4] \( {(1+x^2)y'+2xy ={1 \over (1+x^2)y}}\)

[exer:2.4.5] \(y'-xy=x^3y^3; \quad \{-3\le x\le 3,\-2\le y\ge 2\}\)

[exer:2.4.6] \( {y'-{1+x\over 3x}y=y^4}; \quad \{-2\le x\le2,-2\le y \le2\}\)

[exer:2.4.7] \(y'-2y=xy^3,\quad y(0)=2\sqrt2\)

[exer:2.4.8] \(y'-xy=xy^{3/2},\quad y(1)=4\)

[exer:2.4.9] \(xy'+y=x^4y^4,\quad y(1)=1/2\)

[exer:2.4.10] \(y'-2y=2y^{1/2},\quad y(0)=1\)

[exer:2.4.11] \( {y'-4y={48x\over y^2},\quad y(0)=1}\)

[exer:2.4.12] \(x^2y'+2xy=y^3,\quad y(1)=1/\sqrt2\)

[exer:2.4.13] \(y'-y=xy^{1/2},\quad y(0)=4\)

[exer:2.4.14] You may have noticed that the logistic equation \[P'=aP(1-\alpha P)\] from Verhulst’s model for population growth can be written in Bernoulli form as \[P'-aP=-a\alpha P^2.\] This isn’t particularly interesting, since the logistic equation is separable, and therefore solvable by the method studied in Section 2.2. So let’s consider a more complicated model, where \(a\) is a positive constant and \(\alpha\) is a positive continuous function of \(t\) on \([0,\infty)\). The equation for this model is \[P'-aP=-a\alpha(t) P^2,\] a non-separable Bernoulli equation. Assuming that \(P(0)=P_0>0\), find \(P\) for \(t>0\). Verify that your result reduces to the known results for the Malthusian model where \(\alpha=0\), and the Verhulst model where \(\alpha\) is a nonzero constant. Assuming that \[\lim_{t\to\infty}e^{-at}\int_0^t\alpha(\tau)e^{a\tau}\,d\tau=L\] exists (finite or infinite), find \(\lim_{t\to\infty}P(t)\).

[exer:2.4.15] \(y'= {y+x\over x}\)

[exer:2.4.16] \(y'= {y^2+2xy \over x^2}\)

[exer:2.4.17] \(xy^3y'=y^4+x^4\)

[exer:2.4.18] \(y'= {y\over x}+\sec{y\over x}\)

[exer:2.4.19] \(x^2y'=xy+x^2+y^2; \quad \{-8\le x\le 8,-8\le y\le 8\}\)

[exer:2.4.20] \(xyy'=x^2+2y^2; \quad \{-4\le x\le 4,-4\le y\le 4\}\)

[exer:2.4.21] \(y'= {2y^2+x^2e^{-(y/x)^2}\over 2xy}; \quad \{-8\le x\le 8,-8\le y\le 8\}\)

[exer:2.4.22] \(y'= {xy+y^2\over x^2}, \quad y(-1)=2\)

[exer:2.4.23] \(y'= {x^3+y^3\over xy^2}, \quad y(1)=3\)

[exer:2.4.24] \(xyy'+x^2+y^2=0, \quad y(1)=2\)

[exer:2.4.25] \(y'= {y^2-3xy-5x^2 \over x^2}, \quad y(1)=-1\)

[exer:2.4.26] \(x^2y'=2x^2+y^2+4xy, \quad y(1)=1\)

[exer:2.4.27] \(xyy'=3x^2+4y^2, \quad y(1)=\sqrt{3}\)

[exer:2.4.28] \(y'= {x+y \over x-y}\)

[exer:2.4.29] \((y'x-y)(\ln |y|-\ln |x|)=x\)

[exer:2.4.30] \(y'= {y^3+2xy^2+x^2y+x^3\over x(y+x)^2}\)

[exer:2.4.31] \(y'= {x+2y \over 2x+y}\)

[exer:2.4.32] \(y'= {y \over y-2x}\)

[exer:2.4.33] \(y'= {xy^2+2y^3\over x^3+x^2y+xy^2}\)

[exer:2.4.34] \(y'= {x^3+x^2y+3y^3 \over x^3+3xy^2}\)

[exer:2.4.35]

- Find a solution of the initial value problem \[x^2y'=y^2+xy-4x^2, \quad y(-1)=0 \eqno{\rm(A)}\] on the interval \((-\infty,0)\). Verify that this solution is actually valid on \((-\infty,\infty)\).
- Use Theorem [thmtype:2.3.1} to show that (A) has a unique solution on \((-\infty,0)\).
- Plot a direction field for the differential equation in (A) on a square \[\{-r\le x\le r, -r\le y\le r\},\] where \(r\) is any positive number. Graph the solution you obtained in (a) on this field.
- Graph other solutions of (A) that are defined on \((-\infty,\infty)\).
- Graph other solutions of (A) that are defined only on intervals of the form \((-\infty,a)\), where is a finite positive number.

[exer:2.4.36]

- Solve the equation \[xyy'=x^2-xy+y^2 \eqno{\rm(A)}\] implicitly.
- Plot a direction field for (A) on a square \[\{0\le x\le r,0\le y\le r\}\] where \(r\) is any positive number.
- Let \(K\) be a positive integer. (You may have to try several choices for \(K\).) Graph solutions of the initial value problems \[xyy'=x^2-xy+y^2,\quad y(r/2)={kr\over K},\]
- for \(k=1\), \(2\), …, \(K\). Based on your observations, find conditions on the positive numbers \(x_0\) and \(y_0\) such that the initial value problem \[xyy'=x^2-xy+y^2,\quad y(x_0)=y_0, \eqno{\rm(B)}\] has a unique solution (i) on \((0,\infty)\) or (ii) only on an interval \((a,\infty)\), where \(a>0\)?
- What can you say about the graph of the solution of (B) as \(x\to\infty\)? (Again, assume that \(x_0>0\) and \(y_0>0\).)

[exer:2.4.37]

- Solve the equation \[y'={2y^2-xy+2x^2 \over xy+2x^2} \eqno{\rm(A)}\] implicitly.
- Plot a direction field for (A) on a square \[\{-r\le x\le r,-r\le y\le r\}\] where \(r\) is any positive number. By graphing solutions of (A), determine necessary and sufficient conditions on \((x_0,y_0)\) such that (A) has a solution on (i) \((-\infty,0)\) or (ii) \((0,\infty)\) such that \(y(x_0)=y_0\).

[exer:2.4.38] Follow the instructions of Exercise [exer:2.4.37} for the equation \[y'={xy+x^2+y^2 \over xy}.\]

[exer:2.4.39] Pick any nonlinear homogeneous equation \(y'=q(y/x)\) you like, and plot direction fields on the square \(\{-r\le x\le r,\ -r\le y\le r\}\), where \(r>0\). What happens to the direction field as you vary \(r\)? Why?

[exer:2.4.40] Prove: If \(ad-bc\ne 0\), the equation \[y'={ax+by+\alpha \over cx+dy+\beta}\] can be transformed into the homogeneous nonlinear equation \[{dY \over dX}={aX+bY \over cX+dY}\] by the substitution \(x=X-X_0,\ y=Y-Y_0\), where \(X_0\) and \(Y_0\) are suitably chosen constants.

[exer:2.4.41] \(y'= {-6x+y-3 \over 2x-y-1}\)

[exer:2.4.42] \(y'= {2x+y+1 \over x+2y-4}\)

[exer:2.4.43] \(y'= {-x+3y-14 \over x+y-2}\)

[exer:2.4.44] \(3xy^2y'=y^3+x\)

[exer:2.4.45] \(xyy'=3x^6+6y^2\)

[exer:2.4.46] \(x^3y'=2(y^2+x^2y-x^4)\)

[exer:2.4.47] \(y'=y^2e^{-x}+4y+2e^x\)

[exer:2.4.48] \(y'= {y^2+y\tan x+\tan^2 x\over\sin^2x}\)

[exer:2.4.49] \(x(\ln x)^2y'=-4(\ln x)^2+y\ln x+y^2\)

[exer:2.4.50] \(2x(y+2\sqrt x)y'=(y+\sqrt x)^2\)

[exer:2.4.51] \((y+e^{x^2})y'=2x(y^2+ye^{x^2}+e^{

[exer:2.4.52] Solve the initial value problem \[y'+{2\over x}y={3x^2y^2+6xy+2\over x^2(2xy+3)},\quad y(2)=2.\]

[exer:2.4.53] Solve the initial value problem \[y'+{3\over x}y={3x^4y^2+10x^2y+6\over x^3(2x^2y+5)},\quad y(1)=1.\]

[exer:2.4.54] Prove: If \(y\) is a solution of a homogeneous nonlinear equation \(y'=q(y/x)\), so is \(y_1=y(ax)/a\), where \(a\) is any nonzero constant.

[exer:2.4.55] A *generalized* *Riccati equation* is of the form \[y'=P(x)+Q(x)y+R(x)y^2. \eqno{\rm (A)}\] (If \(R\equiv-1\), (A) is a *Riccati equation*.) Let \(y_1\) be a known solution and \(y\) an arbitrary solution of (A). Let \(z=y-y_1\). Show that \(z\) is a solution of a Bernoulli equation with \(n=2\).

[exer:2.4.56] \(y'=1+x - (1+2x)y+xy^2\); \(y_1=1\)

[exer:2.4.57] \(y'=e^{2x}+(1-2e^x)y+y^2\); \(y_1=e^x\)

[exer:2.4.58] \(xy'=2-x+(2x-2)y-xy^2\); \(y_1=1\)

[exer:2.4.59] \(xy'=x^3+(1-2x^2)y+xy^2\); \(y_1=x\)