- 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 \(y\)
\(\displaystyle ⟹{1\over y}={4Ae^{2x}+2x+1\over4}\)
\(\displaystyle ⟹y={4\over 4Ae^{2x}+2x+1}\)

VIDEO LESSON


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 \(y\)
\(\displaystyle xy=e^{-3x} (Ax+B)+2x-{4\over3}\)
\(\displaystyle y=e^{-3x} \left(A+{B\over x}\right)+2-{4\over3x}\)

VIDEO LESSON


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 \(\displaystyle x=t^{{1\over2}}, t=x^2\)
\(\displaystyle y=Ae^{x^2}+Be^{3x^2}+x^2+{4\over3}\)
VIDEO LESSON