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Suppose that $H''(r)-aH'(r)=0$, $H'(0)=\frac{1}{p^2}$. Find the "special solution" for $$H(0)=\dfrac{H'(0)}{a}\cdot(1+\frac{N}{Y}e^w)$$

Solution:

Since $$\lambda^2-a\lambda=0\text{ has roots }\lambda=0 \text{ and }\lambda=a $$ We have: $$H(r)=C_0+C_1e^{ar},\text{ where $C_0,C_1$ =const.} \\H(0)=C_0+C_1,~~H'(0)=aC_1$$ We have:$$C_1=\frac{1}{ap^2},~ C_0=H(0)-C_1=\frac{1}{ap^2}\cdot\frac{N}{Y}e^w$$ Hence, $$H=\frac{1}{ap^2}\cdot\frac{N}{Y}e^w+\frac{1}{ap^2}e^{ar}$$

We deduce that:$$Hap^2Y=Ne^w+Ye^{ar}$$ $$\bf{\text{Happy New Year!!!}}$$

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