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CEMAT104 · Complete solutions

Long Test

Order & degree, linearity, verifying solutions, eliminating constants, families of curves. Try each item yourself first, then reveal one step at a time.

0/25 got it

Question paper
A

Order & Degree

Identify the order and degree (solve, if applicable).

1
A · Order & Degree

Laplace's equation (PDE)

∂2u∂x2+∂2u∂y2=0\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0
2
A · Order & Degree

Derivative inside a sine

sin⁡ ⁣(dydx)=x\sin\!\left(\frac{dy}{dx}\right)=x
3
A · Order & Degree

Radical in the denominator

3x2d3ydx3+(d2ydx2)2=7y\frac{3x^2}{\sqrt{\dfrac{d^3y}{dx^3}+\left(\dfrac{d^2y}{dx^2}\right)^2}}=7y
4
A · Order & Degree

Derivative inside ln

ln⁡ ⁣(d2ydx2)+dydx=x\ln\!\left(\frac{d^2y}{dx^2}\right)+\frac{dy}{dx}=x
5
A · Order & Degree

Negative power of y''

(d2ydx2)−1+dydx=xd2ydx2+cos⁡x\left(\frac{d^2y}{dx^2}\right)^{-1}+\frac{dy}{dx}=x\frac{d^2y}{dx^2}+\cos x
6
A · Order & Degree

Power of 4/3

[ y+x(d2ydx2)2]4/3=x(dydx)\left[\,y+x\left(\frac{d^2y}{dx^2}\right)^2\right]^{4/3}=x\left(\frac{dy}{dx}\right)
B

Linear or Nonlinear

State the reason for nonlinearity.

7
B · Linear or Nonlinear

y''' + eˣy' + y = sin x

y′′′+exy′+y=sin⁡xy'''+e^{x}y'+y=\sin x
8
B · Linear or Nonlinear

x y'' + y = 7

xd2ydx2+y=7x\frac{d^2y}{dx^2}+y=7
9
B · Linear or Nonlinear

√(y') + 3y = cos x

dydx+3y=cos⁡x\sqrt{\frac{dy}{dx}}+3y=\cos x
10
B · Linear or Nonlinear

y'' + cos(x + y) = eˣ

y′′+cos⁡(x+y)=exy''+\cos(x+y)=e^{x}
11
B · Linear or Nonlinear

y' + (3x² − 1)/(x + y) = 0

dydx+3x2−1x+y=0\frac{dy}{dx}+\frac{3x^2-1}{x+y}=0
12
B · Linear or Nonlinear

(x² + y²)dx + 2xy dy = 0

(x2+y2) dx+2xy dy=0(x^2+y^2)\,dx+2xy\,dy=0
C

Verify the Solutions

Substitute and show LHS = RHS.

13
C · Verify the Solutions

Is x² = 4 − 4y a solution?

Is x2=4−4yx^2=4-4y a particular solution of (y+x2+y2)dx−x dy=0\left(y+\sqrt{x^2+y^2}\right)dx-x\,dy=0?
14
C · Verify the Solutions

2y = x⁵ + cx³

Show that 2y=x5+cx32y=x^5+cx^3 is a solution of (x5+3y) dx−x dy=0(x^5+3y)\,dx-x\,dy=0.
15
C · Verify the Solutions

Is Aeˣ + Bxeˣ a solution?

Is y=Aex+Bxexy=Ae^{x}+Bxe^{x} a solution of y′′+4y′−4y=0y''+4y'-4y=0?

Typo in the paper. Printed as "y′′+4′−4yy''+4'-4y". The missing yy is assumed.

16
C · Verify the Solutions

2y = x² − 1 + 2e^(1−x²)

Show that 2y=x2−1+2e1−x22y=x^2-1+2e^{1-x^2} is a solution of y′=x3−2xyy'=x^3-2xy.
17
C · Verify the Solutions

Implicit ln – tan⁻¹ solution

Show that (x−3y+2) dx+3(x+3y−4) dy=0(x-3y+2)\,dx+3(x+3y-4)\,dy=0 has ln⁡ ⁣[(x−1)2+9(y−1)2]−2tan⁡−1 ⁣[x−13(y−1)]=C\ln\!\left[(x-1)^2+9(y-1)^2\right]-2\tan^{-1}\!\left[\dfrac{x-1}{3(y-1)}\right]=C as a solution.
D

Eliminate the Constants

n constants → differentiate n times.

18
D · Eliminate the Constants

y = Ae⁴ˣ + Bxe⁴ˣ

y=Ae4x+Bxe4xy=Ae^{4x}+Bxe^{4x}
19
D · Eliminate the Constants

x = c₁cos ωt + c₂sin ωt + c₃cos ωt

x=c1cos⁡ωt+c2sin⁡ωt+c3cos⁡ωt(ω is a parameter)x=c_1\cos\omega t+c_2\sin\omega t+c_3\cos\omega t\qquad(\omega\text{ is a parameter})

Typo in the paper. The c3cos⁡ωtc_3\cos\omega t term repeats c1cos⁡ωtc_1\cos\omega t. It is solved as printed.

20
D · Eliminate the Constants

y = e³ˣ(C₁cos 2x + C₂sin 2x)

y=C1e3xcos⁡2x+C2e3xsin⁡2xy=C_1e^{3x}\cos2x+C_2e^{3x}\sin2x
21
D · Eliminate the Constants

y = c₁x² + c₂e⁻²ˣ

y=c1x2+c2e−2xy=c_1x^2+c_2e^{-2x}
E

Families of Curves

Obtain the DE, with graphs.

22
E · Families of Curves

Lines at fixed distance Q from origin

Straight lines at a fixed distance QQ from the origin.

Graph · drag the sliders

xy
x·cos 35° + y·sin 35° = 2

Every line is tangent to the dashed circle of radius Q.

23
E · Families of Curves

Circles of radius r tangent to y-axis

Circles with fixed radius rr and tangent to the yy-axis.

Graph · drag the sliders

xy
(x − 1.3)² + (y − 1)² = 1.69

Same radius; each circle just touches the y-axis.

24
E · Families of Curves

Circles of radius 1

Circles of radius unity.

Graph · drag the sliders

xy
(x − 0.5)² + (y − 0.5)² = 1

The radius is locked at 1. Only the center moves (2 constants, so 2nd order).

25
E · Families of Curves

Sideways parabolas, focal length a

Parabolas with axis parallel to the xx-axis and with distance from vertex to focus fixed as aa.

Graph · drag the sliders

xy
(y − 0.5)² = 2(x + 1)

Same shape (same a), only the vertex moves. The 2nd dot is the focus.