5.7 Differential Equations

Consider the differential equation

`cos^2 x ( dy)/( dx)=1-y,-π/2 < x < π/2`

with `y=2` when `x=0`.

(a) Find the integrating factor for the above differential equation.

(b) Find the solution of the differential equation.

(a) Integrating factor is `e^tanx`.

(b) Multiplying the given equation by the integrating factor, we get,

`y^' e^tanx+ye^tanx sec^2 x=e^tanx sec^2 x`

or: `d/dx (ye^tanx )=e^tanx sec^2 x` 

or: `ye^tanx=∫e^tanx sec^2 x dx=e^tanx+k`

or: `y=1+ke^(-tanx)`

Since `y=2` when `x=0`, we have

`2=1+ke^0 " or " k=1`

The solution over `(-π/2,π/2)` is `y=1+e^(-tanx)`.

Consider the homogeneous differential equation `dy/( dx)=(y^2-2x^2)/(xy)`, where `x,y≠0`.

It is given that `y=2` when `x=1`.

(a) By using the substitution `y=vx`, solve the differential equation. Give your answer in the form `y^2=f(x)`.

The points of zero gradient on the curve `y^2=f(x)` lie on two straight lines of the form `y=mx` where `m∈R`.

(b) Find the values of `m`

(a)  `dy/( dx)=(y^2-2x^2)/(xy)`

let `y=vx`

`⇒dy/( dx)=v+x ( dv)/( dx)`

`⇒v+x ( dv)/( dx)=(v^2 x^2-2x^2)/(vx^2 )`

`⇒x ( dv)/( dx)=(-2)/v`

`⇒∫v dv=-∫2/x dx`

`⇒v^2/2=-2ln|x|(+c)`

`⇒y^2/(2x^2 )=-2ln|x|+c`

attempt to substitute `x=1, y=2` into their integrated expression to find `c`

 `⇒2=-2ln|1|+c⇒c=2`

`⇒y^2/(2x^2 )=-2ln|x|+2`

 `⇒y^2=2x^2 (-2ln|x|+2)(=4x^2 (1-ln|x|))`

(b) attempt to set `(dy)/(dx)=0` in the differential equation

`y=sqrt2 x" and " y=-sqrt2 x" or " m=±sqrt2`

Solve the differential equation

`dy/( dx)=ytanx+1, 0 ≤ x < π/2`

if `y=1` when `x=0`.

 `dy/( dx)=ytanx+1, 0 ≤ x < π/2`

 `dy/( dx)-ytanx=1`

 `h(x)=∫-tanx dx=ln(cosx)`

The integrating factor is `e^(h(x))=cosx.`.

Now,  `cosx ( dy)/( dx)-ysinx=cosx`

`d/dx(ycosx)=cosx`

 `ycosx=∫cosx dx=sinx+c`

 `y=tanx+csecx`

But, `y=1` when `x=0` giving `c=1`.

 `y=tanx+secx`

Answer : `y=tanx+secx` OR `ycosx=sinx+1`

Solve the differential equation `dy/( dx)=(ln2x)/x^2 -(2y)/x,x>0`, given that `y=4` at `x=1/2`. Give your answer in the form `y=f(x)`.

 `dy/( dx)+(2y)/x=(ln2x)/x^2`

attempt to find integrating factor

 `(e^(∫2/( x) dx)=e^(2lnx))=x^2`

 `x^2 ( dy)/( dx)+2xy=ln2x`

 `d/dx (x^2 y)=ln2x`

 `x^2 y=∫ln2x dx`

attempt to use integration by parts

 `x^2 y=xln2x-x(+c)`

`y=(ln2x)/x-1/x+c/x^2`

substituting `x=1/2, y=4` into an integrated equation involving `c`

`4=0-2+4c ⇒c=3/2`

  `y=(ln2x)/x-1/x+3/(2x^2 )` 

Consider the differential equation `dy/( dx)=(4x^2+y^2-xy)/x^2`, with `y=2` when `x=1`.

(a) Use Euler's method, with step length `h=0.1`, to find an approximate value of  `y` when `x=1.4`.

(b) Sketch the isoclines for `(dy)/dx=4`.

(c) (i) Express `m^2-2m+4` in the form `(m-a)^2+b`, where `a,b∈Z`. 

    (ii) Solve the differential equation, for `x>0`, giving your answer in the form `y=f(x)`.

  (iii) Sketch the graph of  `y=f(x)` for `1 ≤ x ≤ 1.4`.

  (iv) With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture `f(1.4)` will be less than, equal to, or greater than your answer in part (a).

(a)

`y(1.4)≈5.34` 

(b) attempt to solve

`(4x^2+y^2-xy)/x^2 =4`

`⇒y^2-xy=0`

`y(y-x)=0`

`y=0" or " y=x`

(c) (i) `m^2-2m+4=(m-1)^2+3 (a=1,b=3)`

(ii) recognition of homogeneous equation, let `y=vx` the equation can be written as

 `v+x( dv)/(dx)=4+v^2-v`

`x ( dv)/( dx)=v^2-2v+4`

`∫1/(v^2-2v+4) dv=∫1/x dx`

`∫1/((v-1)^2+3) dv=∫1/x dx" from part (c)(i) "`

`1/sqrt3 arctan((v-1)/sqrt3)=lnx(+c)`

`x=1,y=2⇒v=2`

`1/sqrt3 arctan(1/sqrt3)=ln1+c`

`⇒c=π/(6sqrt3) (=0.302)`

`arctan((v-1)/sqrt3)=sqrt3 lnx+π/6`

`" substituting " v=y/x`

`y=x(sqrt3 tan(sqrt3 lnx+π/6)+1)`

(iii)

curve drawn over correct domain

(iv) the sketch shows that `f` is concave up

this means the tangent drawn using Euler's method will give an underestimate of the real value, so `f(1.4) >` estimate in part (a)

Consider the differential equation `dy/( dx)=x/e^(2y)`.

(a) 

Identify which of the following diagrams, A,B or C, represents the slope field for the differential equation. Give a reason for your answer.

It is given that, for a particular solution, `x=0` and `y=0`.

(b) Find an expression for `y`, in terms of `x`, for this solution.

(c) Find `(dy)/dx`, in terms of `x`, by differentiating your answer from part (b).

(d) Hence verify that your answer to part (b) is a solution to `(dy)/( dx)=x/e^(2y)`.

(a) C.

Any valid reason for accepting C . or rejecting A . and B .

for example:

• when `x=0` slopes have (or appear to have) zero gradient
• (slope field is) always positive for `x>0`

 (b) `∫e^(2y) dy=∫x dx`

`1/2 e^(2y)=1/2 x^2 (+c)`

substituting in `x=0, y=0`

 `1/2=c`

`e^(2y)=x^2+1`

`y=1/2 ln(x^2+1)`

(c) `dy/( dx)=1/2×2x×1/(x^2+1) (=x/(x^2+1))`

(d) substitution of `e^(2y)=x^2+1` from part (b) into part(c)(i) or original differential equation `dy/( dx)=x/(x^2+1)=x/e^(2y)`

and hence `y=1/2 ln(x^2+1)` is a solution for the differential equation

A particle moves such that its displacement, `x` metres, from a point O at time `t` seconds is given by the differential equation

`(d^2 x)/( dt^2 )+5 ( dx)/( dt)+6x=0`

(a) (i) Use the substitution `y=(dx)/dt` to show that this equation can be written as `(((dx)/dt),((dy)/(dt)))=((0,1),(-6,-5))((x),(y))`

     (ii) Find the eigenvalues for the matrix `((0,1),(-6,-5))`.

    (iii) Hence state the long-term velocity of the particle.

The equation for the motion of the particle is amended to

`(d^2 x)/( dt^2 )+5 ( dx)/( dt)+6x=3t+4`

(b) (i) Use the substitution `y=(dx)/dt` to write the differential equation as a system of coupled, first order differential equations.

When  the particle is stationary at O.

    (ii) Use Euler's method with a step length of 0.1 to find the displacement of the particle when `t=1`.

   (iii) Find the long-term velocity of the particle.

(a) (i) `y=dx/( dt)⇒( dy)/( dt)+5 ( dx)/( dt)+6x=0`

`(((dx)/dt),((dy)/(dt)))=((0,1),(-6,-5))((x),(y))`

(ii) `det((-λ,1),(-6,-5-λ))=0`

`-λ(-5-λ)+6=0`

 `λ^2+5λ+6=0`

   `λ=-2,-3`

(iii) (on a phase portrait the particle approaches `(0,0)` as `t` increases so long term velocity `(y)` is 0

(b) (i)  `y=(dx)/dt`

`(d^2 x)/( dt^2 )=( dy)/( dt)`

`(dy)/dt+5y+6x=3t+4`

(ii) recognition that `h=0.1` in any recurrence formula

`(t_(n+1)=t_n+0.1)`

`x_(n+1)=x_n+0.1y_n`

`y_(n+1)=y_n+0.1(3t_n+4-5y_n-6x_n )`

`(" when " t=1,)x=0.64402…≈0.644 m`

(iii) recognizing that `y` is the velocity

`0.5ms^-1`

The curve `y=f(x)` passes through the point `(2,6)`. Given that `(dy)/dx=3x^3-5`, find `y` in terms of `x`.

Attempting to integrate.

`y=x^3-5x+c`

 substitute `(2,6)` to find `c` (`(6=2^3-5(2)+c)` 

 `c=8`

`y=x^3-5x+8` (Accept `x^3-5x+8`).

When air is released from an inflated balloon it is found that the rate of decrease of the volume of the balloon is proportional to the volume of the balloon. This can be represented by the differential equation `(dv)/dt=-kv`, where `v` is the volume, `t` is the time and `k` is the constant of proportionality.

(a) If the initial volume of the balloon is `v_o`, find an expression, in terms of `k`, for the volume of the balloon at time `t`.

(b) Find an expression, in terms of `k`, for the time when the volume is `v_o/2`.

(a) Given `(dv)/dt=-kv`

`⇔∫(dv)/v=-k∫dt`

`⇔lnv=-kt+C`

`⇔v=Ae^(-kt) (A=e^C )`

At `t=0,v=v_0⇒A=v_0`

`⇔v=v_0 e^(-kt)`

(b) Put `v=v_0/2`

then `v_0/2=v_0 e^(-kt)`

`⇔1/2=e^(-kt)`

`⇔ln 1/2=-kt`

`⇔t=ln2/k`

Consider the differential equation `2xy ( dy)/( dx)=y^2-x^2`, where `x>0`.

(a) 

Solve the differential equation and show that a general solution is `x^2+y^2=cx` where `c` is a positive constant.

(b) Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of `c`.

(a) `dy/( dx)=(y^2-x^2)/(2xy)`

let `y=vx`

`dy/( dx)=v+x ( dv)/( dx)`

`v+x ( dv)/( dx)=(v^2 x^2-x^2)/(2vx^2)`

`v+x ( dv)/( dx)=(v^2-1)/(2v) (=v/2-1/(2v))`

`x ( dv)/( dx)=(-v^2-1)/(2v) (=-v/2-1/(2v))`

 `(2v)/(1+v^2 ) ( dv)/( dx)=-1/x`

 `∫(2v)/(1+v^2 ) dv=∫-1/x dx`

 `ln(1+v^2 )=-lnx+lnc`

`ln(1+(y/x)^2 )=-lnx+lnc`

`1+(y/x)^2=c/x`

`x^2+y^2=cx`

(b)

`x^2+y^2=cx`

`(x-c/2)^2+y^2=c^2/4`

this is a circle radius `c/2` centre `(c/2,0)` 

hence there are two tangents

tangents at `y=c/2, y=-c/2`