Differential EquationshardPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Linear ODE with Radical: ⌊y(√5)⌋ = 3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let y=y(x)y=y(x) be the solution of the differential equation (x2−xx2−1)dy+(y(x−x2−1)−x)dx=0\left(x^2-x\sqrt{x^2-1}\right)dy+\left(y\left(x-\sqrt{x^2-1}\right)-x\right)dx=0, x≥1x\ge1. If y(1)=1y(1)=1, then the greatest integer less than y(5)y\left(\sqrt5\right) is
Solution
Answer: 3 (± 0.01)
Step 1: x2−xx2−1=x(x−x2−1)x^2-x\sqrt{x^2-1}=x\left(x-\sqrt{x^2-1}\right). Divide by x(x−x2−1) dxx\left(x-\sqrt{x^2-1}\right)\,dx:
dydx+yx=xx(x−x2−1)=1x−x2−1.\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{x}{x\left(x-\sqrt{x^2-1}\right)}=\dfrac{1}{x-\sqrt{x^2-1}}.
 Rationalise (×x+x2−1x+x2−1\times\dfrac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}}, denominator x2−(x2−1)=1x^2-(x^2-1)=1):
1x−x2−1=x+x2−1⇒dydx+yx=x+x2−1.\dfrac{1}{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1}\Rightarrow\dfrac{dy}{dx}+\dfrac{y}{x}=x+\sqrt{x^2-1}.
 Step 2: P(x)=1x⇒P(x)=\dfrac1x\Rightarrow I.F. =e∫dx/x=x=e^{\int dx/x}=x:
ddx(xy)=x(x+x2−1)=x2+xx2−1.\dfrac{d}{dx}(xy)=x\left(x+\sqrt{x^2-1}\right)=x^2+x\sqrt{x^2-1}.
 Step 3: ∫x2 dx=x33\displaystyle\int x^2\,dx=\dfrac{x^3}{3}; for ∫xx2−1 dx\displaystyle\int x\sqrt{x^2-1}\,dx, u=x2−1, du=2x dxu=x^2-1,\ du=2x\,dx: ∫u du2=(x2−1)3/23\int\sqrt u\,\tfrac{du}{2}=\dfrac{(x^2-1)^{3/2}}{3}.
⇒xy=x33+(x2−1)3/23+C.\Rightarrow xy=\dfrac{x^3}{3}+\dfrac{(x^2-1)^{3/2}}{3}+C.
 Step 4: y(1)=1y(1)=1: at x=1x=1, x3=1x^3=1, (x2−1)3/2=0(x^2-1)^{3/2}=0, xy=1xy=1:
1=13+0+C⇒C=23.1=\dfrac13+0+C\Rightarrow C=\dfrac23.
 Step 5: x=5x=\sqrt5: x3=55x^3=5\sqrt5, x2−1=4⇒(x2−1)3/2=8x^2-1=4\Rightarrow(x^2-1)^{3/2}=8:
5 y=553+83+23=55+103.\sqrt5\,y=\dfrac{5\sqrt5}{3}+\dfrac{8}{3}+\dfrac23=\dfrac{5\sqrt5+10}{3}.
y(5)=55+1035=53+253≈1.667+1.491=3.157.y(\sqrt5)=\dfrac{5\sqrt5+10}{3\sqrt5}=\dfrac{5}{3}+\dfrac{2\sqrt5}{3}\approx1.667+1.491=3.157.
 Step 6: ∴⌊y(5)⌋=3\therefore\lfloor y(\sqrt5)\rfloor=3. Correct answer: 3
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