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Solve a homogeneous equation by putting y = vx

This type of question, asked 10 times: May 2022 S1, 2 May 2023 S1, 11 May 2023 S1, 23 Apr 2025 S1, 25 Apr 2025 S2, 13 Apr 2026 S1, 17 Apr 2026 S1, 19 Apr 2026 S1, 13 May 2026 S1, 15 May 2026 S2. All 10 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    May 2022 S1

    109
    The general solution of the differential equation dydx=3x+yx−y\frac{dy}{dx} = \frac{\sqrt{3}x+y}{x-y} is (where C is a constant of integration.)

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  2. Question 2

    2 May 2023 S1

    132
    If xdydx=y(log⁡y−log⁡x+1)x\dfrac{dy}{dx} = y(\log y - \log x + 1), then general solution of this equation is

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  3. Question 3

    11 May 2023 S1

    137
    The particular solution of the differential equation xydydx=x2+2y2xy\frac{dy}{dx} = x^2 + 2y^2 when y(1)=0y(1) = 0 is

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  4. Question 4

    23 Apr 2025 S1

    115
    If x dy dx=y(log⁡y−log⁡x+1)x\frac{\text{ }dy}{\text{ }dx}=y(\log y- \log x+ 1), then the solution of the equation is

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  5. Question 5

    25 Apr 2025 S2

    150
    The general solution of differential equation (y2−x2)dx=xy dy(x≠0)\left( y^{2}-x^{2} \right)dx=xy\text{ }dy(x\neq 0) is

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  6. Question 6

    13 Apr 2026 S1

    84
    The solution of the differential equation dydx=x+yx−y\dfrac{dy}{dx} = \dfrac{x + y}{x - y} is

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  7. Question 7

    17 Apr 2026 S1

    88
    For the differential equation (x2+y2) dy=xy dx(x^2 + y^2)\,dy = xy\,dx, it is given that y(1)=1y(1) = 1 and y(x0)=ey(x_0) = e, then the value of x0x_0 is ____

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  8. Question 8

    19 Apr 2026 S1

    86
    The solution of the differential equation dydx=x−yx+y\dfrac{dy}{dx} = \dfrac{x - y}{x + y}, when x=0x = 0 and y=0y = 0 represents ....

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  9. Question 9

    13 May 2026 S1

    86
    The solution of the differential equation y2−x2dydx=xydydxy^2 - x^2\dfrac{dy}{dx} = xy\dfrac{dy}{dx} is ...

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  10. Question 10

    15 May 2026 S2

    86
    If the general solution of the differential equation ysec⁡2(xy)dx−(xsec⁡2(xy)−y)dy=0y\sec^2\left(\dfrac{x}{y}\right)dx - \left(x\sec^2\left(\dfrac{x}{y}\right) - y\right)dy = 0 is tan⁡(xy)+g(y)=c\tan\left(\dfrac{x}{y}\right) + g(y) = c, then g(y)=…g(y) = \ldots

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