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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the differential equation: $(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}-4\text{x}^2+0,$ subject to the initial condition $\text{y}(0)=0.$
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Answer

The given differential equation can be written as:
$\frac{\text{dy}}{\text{dx}}+\frac{2\text{x}}{1+\text{x}^2}\text{y}=\frac{4\text{x}^2}{1+\text{x}^2}\ \dots(1)$
This is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\frac{4\text{x}}{1+\text{x}^2}$
$\text{I.F}=\text{e}^{{\int}\text{Pdx}}=\text{e}^{{\int}\frac{2\text{x}}{1+\text{x}^2}\text{dx}}=\text{e}^{{\log}(1+\text{x}^2)}=1+\text{x}^2$
Multipying both sides of (1) by $\text{I.F}.=(1+\text{x}^2),$ we get
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^2$
Integrating both sides with respect to x, we get
$\text{y}(1+\text{x}^2)=\int4\text{x}^2\text{dx}+\text{C}$
$\text{y}(1+\text{x}^2)=\frac{4\text{x}^3}{3}+\text{C}\ \dots(2)$
Given $\text{y}=0,$ when $\text{x}=0$
Substituting $\text{x}=0$ and $\text{y}=0$ in (1), we get
$0=0+\text{C}\Rightarrow\text{C}=0$
Substituting $\text{C}=0$ in (2), we get $\text{y}=\frac{4\text{x}^3}{3(1+\text{x}^2)},$ which is the required solution.

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