JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 4)
Which of the following is true for y(x) that satisfies the differential equation
$${{dy} \over {dx}}$$ = xy $$-$$ 1 + x $$-$$ y; y(0) = 0 :
$${{dy} \over {dx}}$$ = xy $$-$$ 1 + x $$-$$ y; y(0) = 0 :
y(1) = 1
y(1) = e$$-$$$${1 \over 2}$$ $$-$$ 1
y(1) = e$${1 \over 2}$$ $$-$$ e$$-$$$${1 \over 2}$$
y(1) = e$${1 \over 2}$$ $$-$$ 1
Explanation
$${{dy} \over {dx}} = (x - 1)y + (x - 1)$$
$${{dy} \over {dx}} = (x - 1)(y + 1)$$
$${{dy} \over {y + 1}} = (x - 1)dx$$
Integrating both sides, we get
$$\ln (y + 1) = {{{x^2}} \over 2} - x + c$$
$$x = 0,y = 0$$
$$ \Rightarrow c = 0$$
$$ \therefore $$ $$\ln (y + 1) = {{{x^2}} \over 2} - x$$
putting $$x = 1,\ln (y + 1) = {1 \over 2} - 1 = - {1 \over 2}$$
$$y + 1 = {e^{ - {1 \over 2}}}$$
$$y = {e^{ - {1 \over 2}}} - 1$$
$$ \therefore $$ $$y(1) = {e^{ - {1 \over 2}}} - 1$$
$${{dy} \over {dx}} = (x - 1)(y + 1)$$
$${{dy} \over {y + 1}} = (x - 1)dx$$
Integrating both sides, we get
$$\ln (y + 1) = {{{x^2}} \over 2} - x + c$$
$$x = 0,y = 0$$
$$ \Rightarrow c = 0$$
$$ \therefore $$ $$\ln (y + 1) = {{{x^2}} \over 2} - x$$
putting $$x = 1,\ln (y + 1) = {1 \over 2} - 1 = - {1 \over 2}$$
$$y + 1 = {e^{ - {1 \over 2}}}$$
$$y = {e^{ - {1 \over 2}}} - 1$$
$$ \therefore $$ $$y(1) = {e^{ - {1 \over 2}}} - 1$$
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