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

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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]

1. 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 $$(-\infty,\infty)$$.
2. Use Theorem [thmtype:2.3.1} to show that (A) has a unique solution on $$(-\infty,0)$$.
3. 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.
4. Graph other solutions of (A) that are defined on $$(-\infty,\infty)$$.
5. 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]

1. Solve the equation $xyy'=x^2-xy+y^2 \eqno{\rm(A)}$ implicitly.
2. 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.
3. Let $$K$$ be a positive integer. (You may have to try several choices for $$K$$.) Graph solutions of the initial value problems $xyy'=x^2-xy+y^2,\quad y(r/2)={kr\over K},$
4. 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 $$a>0$$?
5. 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]

1. Solve the equation $y'={2y^2-xy+2x^2 \over xy+2x^2} \eqno{\rm(A)}$ implicitly.
2. 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] $$xy'=x^3+(1-2x^2)y+xy^2$$; $$y_1=x$$