The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{3…
The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{3 x^2}{1+x^3}\right) y=\frac{1}{x^3+1}$ is
- $y\left(1+x^3\right)=x^3+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $y\left(1+x^3\right)=x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $y\left(1+x^3\right)=x^2+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $y\left(1+x^3\right)=2 x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Solution
Given differential equation is
$\begin{array}{ll}
& \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{3 x^2}{1+x^3}\right) y=\frac{1}{x^3+1} \\
& \text { Here, } \mathrm{P}=\frac{3 x^2}{1+x^3}, \mathrm{Q}=\frac{1}{x^3+1} \\
\therefore \quad & \text { I.F. }=\mathrm{e}^{\int \frac{3 x^2}{1+x^3} \mathrm{~d} x}=\mathrm{e}^{\log \left(1+x^3\right)}=\left(1+x^3\right) \\
\therefore \quad & \text { Solution of the given equation is } \\
& y\left(1+x^3\right)=\int \frac{1}{1+x^3} \cdot\left(1+x^3\right) \mathrm{d} x+\mathrm{c} \\
& \Rightarrow y\left(1+x^3\right)=x+\mathrm{c}
\end{array}$
Asked in: MHT CET 2023 (10 May Shift 2)
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