It is given that $y\frac{\mathrm{d}y}{\mathrm{d}x}+x=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Using the s

Question

It is given that $y\frac{\mathrm{d}y}{\mathrm{d}x}+x=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Using the substitution $w={{x}^{2}}+{{y}^{2}}$, show that the differential equation can be transformed to $\frac{\mathrm{d}w}{\mathrm{d}x}=\mathrm{f}\left( w \right)$, where the function $\mathrm{f}\left( w \right)$ is to be found.[2] Hence, given that $y=4$ when $x=3$, solve the differential equation $y\frac{\mathrm{d}y}{\mathrm{d}x}+x=\sqrt{{{x}^{2}}+{{y}^{2}}}$.[3] Solve the differential equation $\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=\frac{1}{x}$, where $x>0$.[4]

Accepted Answer

(a)(i) $2\sqrt{w}$ (ii) The particular solution is $\sqrt{{{x}^{2}}+{{y}^{2}}}=x+2$ (or ${{x}^{2}}+{{y}^{2}}={{\left( x+2 \right)}^{2}}$) (b) $x\ln x+{{C}_{1}}x+D$ (where ${{C}_{1}}=C-1$)

2021 EJC P1 Q5

2021 EJC P1 Q5

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