Related Formula
A linear differential equation of the first order matching (dy)/(dx) + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$ uses an Integrating Factor written as:
I.F. = e∫ P(x) dx$$\text{I.F.} = e^{\int P(x) dx}$$
Core Logic
Rearrange the given differential equation terms to express it in standard linear form:
(1 + x²) (dy)/(dx) + xy = 5x²√(1 + x²)$$(1 + x^2) \frac{dy}{dx} + xy = 5x^2\sqrt{1 + x^2}$$
(dy)/(dx) + ((x)/(1+x²))y = 5x²√(1+x²)$$\frac{dy}{dx} + \left(\frac{x}{1+x^2}\right)y = \frac{5x^2}{\sqrt{1+x^2}}$$
Step 1: Compute Integrating Factor
Calculate the exponent integral for I.F.$\text{I.F.}$:
∫ P(x) dx = ∫ (x)/(1+x²) dx = (1)/(2) ln(1+x²) = ln√(1+x²)$$\int P(x) dx = \int \frac{x}{1+x^2} dx = \frac{1}{2} \ln(1+x^2) = \ln\sqrt{1+x^2}$$
I.F. = eln√(1+x²) = √(1+x²)$$\text{I.F.} = e^{\ln\sqrt{1+x^2}} = \sqrt{1+x^2}$$
Step 2: General Solution and Boundary Condition
The general solution template is:
y · (I.F.) = ∫ Q(x) · (I.F.) dx$$y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx$$
y√(1+x²) = ∫ 5x²√(1+x²) · √(1+x²) dx$$y\sqrt{1+x^2} = \int \frac{5x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} dx$$
y√(1+x²) = ∫ 5x² dx = (5x³)/(3) + C$$y\sqrt{1+x^2} = \int 5x^2 dx = \frac{5x^3}{3} + C$$
Apply the initial condition y(0) = 0$y(0) = 0$:
0 · √(1+0) = 0 + C C = 0$$0 \cdot \sqrt{1+0} = 0 + C \implies C = 0$$
Thus, the explicit functional equation is:
y = 5x³3√(1+x²)$$y = \frac{5x^3}{3\sqrt{1+x^2}}$$
Step 3: Evaluate at Target Value
Substitute x = √(3)$x = \sqrt{3}$ into the isolated function:
y(√(3)) = 5(√(3))³3 1+(√(3))² = 5(3√(3))3√(1+3) = 15√(3)3 · 2 = 5√(3)2$$y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+(\sqrt{3})^2}} = \frac{5(3\sqrt{3})}{3\sqrt{1+3}} = \frac{15\sqrt{3}}{3 \cdot 2} = \frac{5\sqrt{3}}{2}$$
Pattern Recognition
Spotting that multiplying across the differential equation format by √(1+x²)$\sqrt{1+x^2}$ converts the left hand side into a direct product rule derivative matching (d)/(dx)(y√(1+x²))$\frac{d}{dx}(y\sqrt{1+x^2})$ yields a direct integration pathway.
Chapter Mix
Class 12 Mathematics: Differential Equations
