Start learning →

PUCET (Punjab) Homogeneous DE — practice questions

Explore 4 free Homogeneous DE questions in PUCET (Punjab) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

Practise Homogeneous DE with Quill →

Homogeneous DE questions and exercises with answers

Showing questions 1–4 of 4.

To solve a **homogeneous** differential equation of degree 1 (e.g. $dy/dx = f(y/x)$), the standard substitution is:Read the answer and worked solution →Solve the homogeneous DE $x\,dy/dx = x\,\tan(y/x) + y$:Read the answer and worked solution →Solve $\dfrac{dy}{dx} = \dfrac{y + \sqrt{x^2 + y^2}}{x}$.Read the answer and worked solution →Solve $(x^2 - y^2)\,dx + 2xy\,dy = 0$.Read the answer and worked solution →

Worked practice questions

To solve a **homogeneous** differential equation of degree 1 (e.g. $dy/dx = f(y/x)$), the standard substitution is:

  1. $y = x + v$
  2. $y = vx$
  3. $y = v + 1/x$
  4. $y = e^x v$
Show answer and solution

Correct answer: B. $y = vx$

Substitute $y = vx$, so $dy/dx = v + x\,dv/dx$. This converts the equation to a variables-separable form in $v$ and $x$.

Open this question →

Solve the homogeneous DE $x\,dy/dx = x\,\tan(y/x) + y$:

  1. $\sin(y/x) = cx$
  2. $\cos(y/x) = cx$
  3. $\tan(y/x) = cx$
  4. $y/x = cx$
Show answer and solution

Correct answer: A. $\sin(y/x) = cx$

Substitute $y = vx$ so $dy/dx = v + x\,dv/dx$. **Original:** $x(v + x\,dv/dx) = x\tan v + vx$ ⇒ $x\,dv/dx = \tan v$ ⇒ $\cot v\,dv = dx/x$. **Integrating:** $\log\sin v = \log x + \log c$ ⇒ $\sin v = cx$ ⇒ $\sin(y/x) = cx$.

Open this question →

Solve $\dfrac{dy}{dx} = \dfrac{y + \sqrt{x^2 + y^2}}{x}$.

  1. $y + \sqrt{x^2 + y^2} = cx^2$
  2. $y^2 = cx$
  3. $\sqrt{x^2 + y^2} = c$
  4. $y = x + c$
Show answer and solution

Correct answer: A. $y + \sqrt{x^2 + y^2} = cx^2$

**Homogeneous:** put $y = vx$. **After substitution:** $x\,dv/dx = \sqrt{1+v^2}$ ⇒ $dv/\sqrt{1+v^2} = dx/x$. **Integrating:** $\log(v + \sqrt{1+v^2}) = \log(cx)$ ⇒ $v + \sqrt{1+v^2} = cx$. Back-substitute $v = y/x$: $y/x + \sqrt{1 + y^2/x^2} = cx$, which simplifies to $y + \sqrt{x^2+y^2} = cx^2$.

Open this question →

How to practise

  1. Choose a Homogeneous DE question and try to work out your answer.
  2. Open the question to check the correct answer and read its worked solution.
  3. Continue in the app for bite-sized lessons, explanations and review with Quill. You can start without creating an account.

More Mathematics topics

SequencesProbability & StatisticsBinomial TheoremSequences and SeriesGeometryLinear ProgrammingLimits and ContinuityPermutations and Combinations