## - Differential Equations Involving a Substitution

LESSON 1

(i) Show that by using the substitution $$\displaystyle y={1\over z}$$, the differential equation
$$\displaystyle {dy\over dx}+2y=xy^2$$
may be written in the form
$$\displaystyle {dz\over dx}-2z=-x$$
(ii) Find the general solution of
$$\displaystyle {dz\over dx}-2z=-x$$
and hence find the general solution of
$$\displaystyle {dy\over dx}+2y=xy^2$$
SOLUTION
(i) Differentiate implicitly
$$\displaystyle y={1\over z}$$
$$\displaystyle {dy\over dx}={dy\over dz}×{dz\over dx}$$
$$\displaystyle =-{1\over z^2} ×{dz\over dx}$$

• Substitute expressions for $$y$$ and $$\displaystyle {dy\over dx}$$ into equation
$$\displaystyle {dy\over dx}+2y=xy^2$$
$$\displaystyle ⟹\left(-{1\over z^2}\right) \left({dz\over dx}\right)+2\left({1\over z}\right)=x\left({1\over z}\right)^2$$
Multiply by $$-z^2$$
$$\displaystyle ⟹{dz\over dx}-2z=-x$$
(ii) Using $$\displaystyle {dz\over dx}-2z=-x$$
Auxiliary equation:
$$u-2=0$$
$$u=2$$
COMPLEMENTARY FUNCTION
$$z=Ae^{2x}$$
PARTICULAR INTEGRAL
Let $$z=Bx+C$$
$$\displaystyle ⟹{dz\over dx}=B$$
• Substitute into equation $$\displaystyle {dz\over dx}-2z=-x$$
$$⟹B-2(Bx+C)=-x$$
$$⟹B-2C-2Bx=-x$$
• Equate coefficients of $$x$$
$$⟹-2B=-1$$
$$\displaystyle ⟹B={1\over 2}$$
• Equate constants
$$\displaystyle {1\over 2}-2C=0$$
$$\displaystyle {1\over 4}=C$$
PARTICULAR INTEGRAL:
$$\displaystyle z={1\over 2} x+{1\over4}$$

GENERAL SOLUTION
$$z=Ae^{2x}+{1\over2} x+{1\over4}$$
This general solution is written in terms of $$z$$. We need to rewrite in terms of $$y$$.
$$\displaystyle y={1\over z}$$
$$\displaystyle ⟹z={1\over y}$$
• Substitute into general solution
$$\displaystyle ⟹{1\over y}=Ae^{2x}+{1\over2} x+{1\over4}$$
• Solve for $$y$$
$$\displaystyle ⟹{1\over y}={4Ae^{2x}+2x+1\over4}$$
$$\displaystyle ⟹y={4\over 4Ae^{2x}+2x+1}$$

LESSON 2
It is given that $$x≠0, y$$ satisfies the differential equation
$$\displaystyle x {d^2 y\over dx^2}+2(3x+1) {dy\over dx}+3y(3x+2)=18x$$
(a) Show that the substitution $$u=xy$$ transforms this differential equation into
$$\displaystyle {d^2 u\over dx^2}+6 {du\over dx}+9u=18x$$
(b) Hence find the general solution of the differential equation
$$\displaystyle x {d^2 y\over dx^2}+2(3x+1) {dy\over dx}+3y(3x+2)=18x$$
giving your answer in the form $$y=f(x)$$.
SOLUTION
(a) $$u=xy$$
• Differentiate implicitly using the product rule to determine the first and second derivatives.
$$\displaystyle {du\over dx}=y+x {dy\over dx}$$
$$\displaystyle {d^2 u\over dx^2}={dy\over dx}+{dy\over dx}+x {d^2 y\over dx^2}$$
$$\displaystyle =2 {dy\over dx}+x {d^2 y\over dx^2}$$
• Substitute expressions for the first and second derivatives into
$$\displaystyle x {d^2 y\over dx^2}+2(3x+1) {dy\over dx}+3y(3x+2)=18x$$
$$\displaystyle ⟹x {d^2 y\over dx^2}+6x {dy\over dx}+2 {dy\over dx}+9xy+6y=18x$$
$$\displaystyle ⟹\left(x {d^2 y\over dx^2}+2 {dy\over dx}\right)+6\left(y+x {dy\over dx}\right)+9xy=18x$$
$$\displaystyle ⟹{d^2 u\over dx^2}+6 {du\over dx}+9u=18x$$
Auxiliary equation
$$m^2+6m+9=0$$
$$m=-3,-3$$
COMPLEMENTARY FUNCTION
$$u=e^{-3x} (Ax+B)$$
PARTICULAR INTEGRAL
Let $$u=Cx+D$$
$$\displaystyle ⟹{du\over dx}=C$$
$$\displaystyle ⟹{d^2 u\over dx^2}=0$$
• Substitute into
$$\displaystyle {d^2 u\over dx^2}+6 {du\over dx}+9u=18x$$
$$⟹0+6C+9(Cx+D)=18x$$
$$⟹6C+9D+9Cx=18x$$
• Equate coefficients of $$x$$
$$9C=18$$
$$C=2$$
• Equate constants
$$6C+9D=0$$
$$9D=-12$$
$$\displaystyle D=-{4\over3}$$

PARTICULAR INTEGRAL
$$\displaystyle u=2x-{4\over3}$$
GENERAL SOLUTION
$$\displaystyle u=e^{-3x}(Ax+B)+2x-{4\over3}$$
• Rewrite general solution in terms of $$y$$
$$\displaystyle xy=e^{-3x} (Ax+B)+2x-{4\over3}$$
$$\displaystyle y=e^{-3x} \left(A+{B\over x}\right)+2-{4\over3x}$$

LESSON 3
(a) Given that $$\displaystyle x=t^{{1\over2}}, x>0, t>0$$ and $$y$$ is a function of $$x$$, show that:
(i) $$\displaystyle {dy\over dx}=2t^{{1\over2}} {dy\over dt}$$
(ii) $$\displaystyle {d^2 y\over dx^2}=4t {d^2 y\over dt^2}+2 {dy\over dt}$$
(b) Hence show that the substitution $$\displaystyle x=t^{{1\over2}}$$ transforms the differential equation
$$\displaystyle x {d^2 y\over dx^2}-(8x^2+1) {dy\over dx}+12x^3 y=12x^5$$
into
$$\displaystyle {d^2 y\over dt^2}-4 {dy\over dt}+3y=3t$$
(c) Hence find the general solution of the differential equation
$$\displaystyle x {d^2 y\over dx^2}-(8x^2+1) {dy\over dx}+12x^3 y=12x^5$$
giving your answer in the form $$y=f(x)$$.
SOLUTION
(i) $$\displaystyle x=t^{{1\over2}}$$
$$\displaystyle {dy\over dx}={dy\over dt}×{dt\over dx}$$
Now,
$$\displaystyle {dx\over dt}={1\over 2} t^{-{1\over2}}={1\over 2t^{{1\over2}}}$$
$$\displaystyle ⟹{dt\over dx}=2t^{{1\over2}}$$
$$\displaystyle ⟹{dy\over dx}=2t^{{1\over2}} {dy\over dt}$$
• Differentiate implicitly
$$\displaystyle {d\over dx} \left({dy\over dx}\right)={d\over dx} \left(2t^{{1\over2}} {dy\over dt}\right)$$
$$\displaystyle ⟹{d\over dx} \left({dy\over dx}\right)={d\over dt} \left(2t^{{1\over2}} {dy\over dt}\right) {dt\over dx}$$
• Use the product rule
$$\displaystyle ⟹{d^2 y\over dx^2}=\left(t^{-{1\over2}} {dy\over dt}+2t^{1\over2} {d^2 y\over dt^2}\right) {dt\over dx}$$
$$\displaystyle =\left(t^{-{1\over2}} {dy\over dt}+2t^{1\over 2} {d^2 y\over dt^2}\right)\left(2t^{1\over2}\right)$$
• Use properties of indices
$$\displaystyle =2 {dy\over dt}+4t {d^2 y\over dt^2}$$
• Substitute $$\displaystyle x=t^{1\over2}$$ into
$$x {d^2 y\over dx^2}-(8x^2+1) {dy\over dx}+12x^3 y=12x^5$$
• Divide throughout by $$x$$
$$\displaystyle ⟹{d^2 y\over dx^2}-\left(8x+{1\over x}\right) {dy\over dx}+12x^2 y=12x^4$$
• Substitute $$\displaystyle x=t^{{1\over2}}, {dy\over dx}=2t^{{1\over2}} {dy\over dt}$$ and $$\displaystyle {d^2 y\over dx^2}=4t {d^2 y\over dx^2}+2 {dy\over dt}$$
$$\displaystyle ⟹4t {d^2 y\over dt^2}+2 {dy\over dt}-\left(8t^{1\over 2}\right)+t^{-{1\over 2}}\left(2t^{{1\over2}} {dy\over dt}+12\left(t^{{1\over2}}\right) \right)^2 y=12\left(t^{{1\over2}} \right)^4$$
• Simplify using indices and factorization.
$$\displaystyle ⟹4t {d^2 y\over dt^2}+(2-16t-2) {dy\over dt}+12ty=12t^2$$
$$\displaystyle ⟹4t {d^2 y\over dt^2}-16t {dy\over dt}+12ty=12t^2$$
$$\displaystyle ⟹{d^2 y\over dt^2}-4 {dy\over dt}+3y=3t$$
Auxiliary equation
$$u^2-4u+3=0$$
$$u=1, 3$$
COMPLEMENTARY FUNCTION:
$$y=Ae^t+Be^{3t}$$
PARTICULAR INTEGRAL
$$y=Ct+D$$
$$\displaystyle ⟹{dy\over dt}=C$$
$$\displaystyle ⟹{d^2 y\over dt^2}=0$$
• Substitute into equation
$$\displaystyle {d^2 y\over dt^2}-4 {dy\over dt}+3y=3t$$
$$0-4C+3(Ct+D)=3t$$
$$-4C+3Ct+3D=3t$$
• Equate coefficients of $$t$$
$$3C=3$$
$$C=1$$
• Equate constants
$$-4C+3D=0$$
$$-4+3D=0$$
$$3D=4$$
$$\displaystyle D={4\over3}$$
PARTICULAR INTEGRAL
$$\displaystyle y=t+{4\over3}$$
GENERAL SOLUTION
$$\displaystyle y=Ae^t+Be^{3t}+t+{4\over3}$$
• Rewrite in terms of $$x$$
Since $$\displaystyle x=t^{{1\over2}}, t=x^2$$
$$\displaystyle y=Ae^{x^2}+Be^{3x^2}+x^2+{4\over3}$$