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Solve a linear equation using a limit condition, or find a limit of the solution

This type of question, asked 6 times: 2021 Paper 18, 26 Jul 2022, 29 Jul 2022, 25 Jan 2023, 1 Feb 2024, 9 Apr 2024. All 6 questions below, oldest first.

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  1. Question 1

    2021 Paper 18

    64
    Let y=y(x)y=y(x) be a solution curve of the differential equation (y+1)tan⁡2xdx+tan⁡xdy+ydx=0(y+ 1)\tan^{2}xdx+ \tan xdy+ydx= 0, x∈(0,π2)x\in\left( 0,\frac{\pi}{2} \right). If lim⁡x→0+xy(x)=1\lim_{x\rightarrow 0 +} xy(x) = 1, then the value of y(π4)y\left( \frac{\pi}{4} \right) is:

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  2. Question 2

    26 Jul 2022

    178
    Suppose y=y(x)y = y(x) be the solution curve to the differential equation dydx−y=2−e−x\frac{dy}{dx}- y = 2 -e^{- x} such that lim⁡x→∞y(x)\lim_{x \rightarrow \infty} y(x) is finite. If aa and bb are respectively the x−x - and yy-intercepts of the tangent to the curve at x=0x = 0, then the value of a−4ba - 4b is equal to

  3. Question 3

    29 Jul 2022

    67
    Let the solution curve y=y(x)y = y(x) of the differential equation (1+e2x)(dydx+y)=1\left( 1 +e^{2x} \right)\left( \frac{dy}{dx}+ y \right)= 1 pass through the point (0,π2)\left( 0,\frac{\pi}{2} \right). Then, lim⁡x→∞exy(x)\lim_{x \rightarrow \infty} e^{x}y(x) is equal to:

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  4. Question 4

    25 Jan 2023

    165
    Let y=y(t)y = y(t) be a solution of the differential equation dydt+αy=γe−βt\frac{dy}{dt}+ \alpha y = \gamma e^{- \beta t} where, α>0,β>0\alpha > 0,\beta > 0 and γ>0\gamma > 0. Then lim⁡t→∞y(t)\lim_{t \rightarrow \infty} y(t)

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  5. Question 5

    1 Feb 2024

    157
    Let α\alpha be a non-zero real number. Suppose f:R→f:R \rightarrow RR is a differentiable function such that f(0)=2f(0) = 2 and lim⁡x→−∞f(x)=1\lim_{x \rightarrow - \infty} f(x) = 1. If f′(x)=αf(x)+3f^{'}(x) = \alpha f(x) + 3, for all x∈Rx \in R, then f(−log⁡e2)f\left( -\log_{e}2 \right) is equal to

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  6. Question 6

    9 Apr 2024

    173
    For a differentiable function f:IR→IRf:IR \rightarrow IR, suppose f′(x)=3f(x)+αf^{'}(x) = 3f(x) + \alpha, where α∈IR,f(0)=1\alpha \in IR,f(0) = 1 and lim⁡x→−∞f(x)=7\lim_{x \rightarrow - \infty} f(x) = 7. Then 9f(−log⁡e3)9f\left( -\log_{e}3 \right) is equal to