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Equation linear in a function of y (substitute u = g(y))

This type of question, asked 7 times: 2021 Paper 19, 2021 Paper 25, 6 Apr 2023, 31 Jan 2024, 8 Apr 2024, 4 Apr 2025, 28 Jan 2026 S1. All 7 questions below, oldest first.

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  1. Question 1

    2021 Paper 19

    66
    Let y(x)y(x) be the solution of the differential equation 2x2dy+(ey−2x)dx=0,x>02x^{2}dy+\left( e^{y}- 2x \right)dx= 0,x> 0. If y(e)=1y(e) = 1, then y(1)y(1) is equal to:

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  2. Question 2

    2021 Paper 25

    84
    If y=y(x)y=y(x) is the solution of the equation esin⁡ycos⁡ydydx+esin⁡ycos⁡x=cos⁡x,y(0)=0e^{\sin y}\cos y\frac{dy}{dx}+e^{\sin y}\cos x= \cos x,y(0) = 0; then 1+y(π6)+32y(π3)+12y(π4)1 +y\left( \frac{\pi}{6} \right)+\frac{\sqrt{3}}{2}y\left( \frac{\pi}{3} \right)+\frac{1}{\sqrt{2}}y\left( \frac{\pi}{4} \right) is equal to

  3. Question 3

    6 Apr 2023

    156
    If the solution curve f(x,y)=0f(x,y) = 0 of the differential equation (1+log⁡ex)dxdy−xlog⁡ex=ey,x>0\left( 1 +\log_{e}x \right)\frac{dx}{dy}- x\log_{e}x =e^{y},x > 0, passes through the points (1,0)(1,0) and (α,2)(\alpha,2) then αα\alpha^{\alpha} is equal to

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  4. Question 4

    31 Jan 2024

    179
    Let y=y(x)y = y(x) be the solution of the differential equation
    sec⁡2xdx+(e2ytan⁡2x+tan⁡x)dy=00<x<π2,y(π4)=0. If y(π6)=α\begin{matrix} & \sec^{2}xdx +\left( e^{2y}\tan^{2}x + \tan x \right)dy = 0 \\ & 0 < x <\frac{\pi}{2},y\left( \frac{\pi}{4} \right)= 0.\text{~If~}y\left( \frac{\pi}{6} \right)= \alpha \end{matrix}
    Then e8αe^{8\alpha} is equal to

  5. Question 5

    8 Apr 2024

    160
    Let y=y(x)y = y(x) be the solution curve of the differential equation secy dydx+2xsiny=x3cos⁡y\frac{dy}{dx}+ 2xsiny =x^{3}\cos y y(1)=0y(1) = 0. Then y(3)y(\sqrt{3}) is equal to:

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  6. Question 6

    4 Apr 2025

    141
    If a curve y=y(x)y = y(x) passes through the point (1,π2)\left( 1,\frac{\pi}{2} \right) and satisfies the differential equation (7x4cot⁡y−excsc⁡y)dxdy=x5,x≥1\left( 7x^{4}\cot y-e^{x}\csc y \right)\frac{dx}{dy}=x^{5},x\geq 1, then at x=2x= 2, the value of cosy is:

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  7. Question 7

    28 Jan 2026 S1

    70
    Let y=y(x)y=y(x) be the solution of the differential equation xdydx−sin⁡2y=x3(2−x3)cos⁡2y,x≠0.x\frac{dy}{dx}- \sin{2y}=x^{3}\left( 2 -x^{3} \right)\cos^{2}y,x\neq 0. If y(2)=0y(2) =0, then tan⁡(y(1))\tan(y(1)) is equal to

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