Differential Equations
Bernoulli equation
Grade Class 12
Question:
<p>If \(\dfrac{dy}{dx} + y = y^2(\tan x + \sec^2 x)\), \(y(0)=1\). Find the solution set.</p>
<span>\(y=-e^x\cos x\)</span>
<span>\(y=\frac{e^x}{e^x-\sin x}\)</span>
<span>\(y=\frac{1}{1-e^{-x}\sin x}\)</span>
<span>\(y=e^x(\cos x-\sin x)\)</span>
Step-by-Step Solution
Key Concept: Bernoulli equation: let v = 1/y to convert to linear ODE.
<div class='solution'><p><strong>Step 1:</strong> Bernoulli: divide by $y^2$: $y^{-2}\dfrac{dy}{dx} + y^{-1} = \tan x + \sec^2 x$.</p>
<p>Let $v = y^{-1} = 1/y$: $-\dfrac{dv}{dx} + v = \tan x + \sec^2 x$, i.e. $\dfrac{dv}{dx} - v = -(\tan x + \sec^2 x)$.</p>
<p><strong>Step 2:</strong> Linear in $v$: IF $= e^{-x}$.</p>
<p>$\dfrac{d(ve^{-x})}{dx} = -e^{-x}(\tan x + \sec^2 x)$.</p>
<p>$ve^{-x} = -\int e^{-x}\tan x\,dx - \int e^{-x}\sec^2 x\,dx$.</p>
<p>By parts on $\int e^{-x}\sec^2 x\,dx$: $= -e^{-x}\tan x - \int e^{-x}\tan x\,dx$. So the two integrals cancel each other's parts and $ve^{-x} = e^{-x}\tan x + C$, giving $v = \tan x + Ce^x$.</p>
<p><strong>Step 3:</strong> $y(0)=1 \implies v(0)=1$: $1 = 0 + C \implies C = 1$. So $\dfrac{1}{y} = \tan x + e^x$, giving $y = \dfrac{1}{e^x + \tan x} = \dfrac{e^{-x}}{1 + e^{-x}\tan x}$. <strong>Answer: (B)</strong>.</p>
<p class='key-concept'>🔑 Key Concept: Bernoulli equation: $y' + P(x)y = Q(x)y^n$. Let $v = y^{1-n}$ to reduce to linear.</p></div>
Correct Answer: 2