It is given that $x\frac{\mathrm{d}y}{\mathrm{d}x}={{x}^{3}}+xy+2y$, $x\ne 0$. Using the su

Question

It is given that $x\frac{\mathrm{d}y}{\mathrm{d}x}={{x}^{3}}+xy+2y$, $x\ne 0$. Using the substitution $y=v{{x}^{2}}$, show that the differential equation can be reduced to $\frac{\mathrm{d}v}{\mathrm{d}x}=v+1$.[2] Given that $y=1$ when $x=1$, solve the differential equation $x\frac{\mathrm{d}y}{\mathrm{d}x}={{x}^{3}}+xy+2y$.[3]

Accepted Answer

(ii) ${1+\frac{y}{x^2}=2\mathrm{e}^{x-1}}$ or ${\ln\left\vert{}1+\frac{y}{x^2}\right\vert{}=x+\ln 2-1}$

2023 MI P2 Q1

2023 MI P2 Q1

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