Multiply both sides by y and by dx:

Differentiate: dydx = 12(x^2 + 9)^-1/2· 2x = xsqrt(x^2 + 9) = xy, which matches the original equation, and y(0) = sqrt(9) = 3, confirming the condition.

Solve dydx = cos x with y(0) = 2. [2 marks]

Solve dydx = 2xy with y(0) = 1, y > 0. [3 marks]

A tank's water height satisfies dhdt = -3 with h(0) = 12. When is the tank empty? [2 marks]

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VICSpecialist MathematicsQuick questions

Unit 4: Calculus

Quick questions on Differential equations: VCE Specialist Mathematics Unit 4

Q: What is apply the initial condition?

Substitute x = 0, y = 3: 92 = 0 + c, so c = 92. Thus

Q: What is solve for y?

Since y > 0 we take the positive root:

7short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is separate?

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Multiply both sides by

y

and by

\mathrm{d}x

What is apply the initial condition?

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Substitute

x = 0

y = 3

\dfrac{9}{2} = 0 + c

, so

c = \dfrac{9}{2}

. Thus

What is solve for

y

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Since

y > 0

we take the positive root:

What is check?

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Differentiate:

\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}(x^2 + 9)^{-1/2}\cdot 2x = \dfrac{x}{\sqrt{x^2 + 9}} = \dfrac{x}{y}

, which matches the original equation, and

y(0) = \sqrt{9} = 3

, confirming the condition.

What is q1?

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Solve

\frac{\mathrm{d}y}{\mathrm{d}x} = \cos x

with

y(0) = 2

. [2 marks]

What is q2?

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Solve

\frac{\mathrm{d}y}{\mathrm{d}x} = 2xy

with

y(0) = 1

y > 0

. [3 marks]

What is q3?

Show answer

A tank's water height satisfies

\frac{\mathrm{d}h}{\mathrm{d}t} = -3

with

h(0) = 12

. When is the tank empty? [2 marks]

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