question on force of interest with investment at 2 different times

Question

You invested $500$ on Jan $1$ $2012$. To save for this amount, you invest $x$ on Jan $1$ $2008$ and $2x$ on July $1$ $2008$. The force of interest is $\delta_t=0.02t$ where $t$ is $0$ on Jan $1$ $2008$. Find $x$.

The accumulated value at $t_{\frac{1}{2}}=A(\frac{1}{2})=x(1+i)^{\frac{1}{2}}+2x$

$=x({e^{(\delta_t)}}^{0.5}+2x)(e^{\int^4_0 0.002t dt})$

This simplifies to

$500=(xe^{\sqrt{0.08}}+2x)(e^{0.16}-1)$

$\therefore x=866.17$

But the answer is $x=142.26$.

Answer 1

The accumulation function is $$a(t)=a(t_0)\mathrm e^{\int_{t_0}^t\delta(s)\mathrm d s}=a(t_0)\mathrm e^{0.01(t^2-t_0^2)}$$ So the equation of value is $$ a(t_0)\mathrm e^{0.01(t^2-t_0^2)}+a(t_1)\mathrm e^{0.01(t^2-t_1^2)}=500 $$ For $t_0=0$ we have $a(t_0)=x$ and $t_1=\frac{1}{2}$ we have $a(t_1)=2x$ and then $$ x\left[\mathrm e^{0.01(4^2)}+2\mathrm e^{0.01(4^2-0.5^2)}\right]=500 $$ and then $$x=\frac{500}{3.514672}=142.26$$

Answer 2

Hint: The equation is

$$ \large{ \left( x\cdot e^{\int_0^{0.5} 0.02\cdot t \, dt}+2x\right)\cdot e^{\int_{0.5}^4 0.02\cdot t \, dt}=500 }$$