2 marksMathematicsDifference and differential equationsFirst-order differential equationsHardabout 3 min

IIT JAM 2021 Question 54: Difference and differential equations

Let $y(x) > 0$ solve $\dfrac{dy}{dx} + y = y^2$ . If $y(\ln 2) = \tfrac13$ , then $y(\ln 3)$ equals _______ (round off to 2 decimal places).

Answer and worked solution

Answer: 0.25

Separating variables, $\dfrac{dy}{y^2 - y} = dx$ , and partial fractions give $\ln\left|\dfrac{y-1}{y}\right| = x + C$ .

Applying $y(\ln 2) = \tfrac13$ gives $\dfrac{y-1}{y} = -e^x$ , i.e. $y = \dfrac{1}{1 + e^x}$ . At $x = \ln 3$ , $y = \dfrac{1}{1 + 3} = 0.25$ . The answer is $0.25$ .

Plain-text version

Let y(x) > 0 solve dy/dx + y = y². If y(ln 2) = 1/3, then y(ln 3) equals _ (round off to 2 decimal places).

Also written as: Let y(x) > 0 solve dy/dx + y = y^2. If y(ln 2) = 1/3, then y(ln 3) equals _ (round off to 2 decimal places).

Difference and differential equations has come up 20 times in 5 exams since 2014.

What this question tests

This question is on First-order differential equations, part of Difference and differential equations in Mathematics.

Solving equations in a function and its first derivative.

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