<p>Rearrange
<span class="math-container">$$\frac{-1}{d_2-d_3} + \frac{\beta}{d_3} = 0$$</span>
to
<span class="math-container">$$\frac{\beta}{d_3} = \frac{1}{d_2-d_3}$$</span>
Flip it:
<span class="math-container">$$\frac{d_3}{\beta} = d_2-d_3$$</span>
The equation is now linear in <span class="math-container">$d_3$</span>.</p>

CC BY-SA 4.0
2022-03-21T08:10:54.990Z
2022-03-21T08:10:54.990Z
Giskard
<p>This is the first-order condition of a dynamic programming problem where I am trying to get the Euler equation from a sequential problem.</p>
<p>(1) <span class="math-container">$$\frac{\partial V(d_2)}{\partial d_3} = \frac{-1}{d_2-d_3} + \frac{\beta}{d_3} = 0$$</span></p>
<p>(2)<span class="math-container">$$d_3 = \frac{\beta d_2}{1+\beta}$$</span></p>
<p>Can anyone plz show the steps of this simplification from stage (1) to stage (2)?</p>

CC BY-SA 4.0
2022-03-19T20:50:41.670Z
Have you actually tried solving the FOC for $d_3$?
Michael Greinecker
CC BY-SA 4.0
2022-03-20T12:21:00.560Z
Since the OP took the time to rewrite the question from a picture into math-setting I assume the OP has also tried the algebra and is desperate. Hint: Try multiplying with $d_2-d_3$ ...
Jesper Hybel
CC BY-SA 4.0
2022-03-20T13:49:50.743Z
@JesperHybel That wasn't OP.
Michael Greinecker
CC BY-SA 4.0
2022-03-20T17:04:33.853Z
Haha ok my bad ....
Jesper Hybel
<macroeconomics><nonexpert><dynamic-programming>
simplification of FOC
econ_confused
# simplification of FOC

(Asked by: username_0 on 2022-03-19)

This is the first-order condition of a dynamic programming problem where I am trying to get the Euler equation from a sequential problem.

(1) $$\frac{\partial V(d\_2)}{\partial d\_3} = \frac{-1}{d\_2-d\_3} + \frac{\beta}{d\_3} = 0$$

(2)$$d\_3 = \frac{\beta d\_2}{1+\beta}$$

Can anyone plz show the steps of this simplification from stage (1) to stage (2)?

**Tags:** macroeconomics, nonexpert, dynamic-programming

> **Comments:**
> **username_1**: Have you actually tried solving the FOC for $d_3$?  
> **username_2**: Since the OP took the time to rewrite the question from a picture into math-setting I assume the OP has also tried the algebra and is desperate. Hint: Try multiplying with $d_2-d_3$ ...  
> **username_1**: @username_2 That wasn't OP.  
> **username_2**: Haha ok my bad ....  

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## 1 answers

### Answer by username_3 on 2022-03-21. Score: 3

Rearrange
$$\frac{-1}{d\_2-d\_3} + \frac{\beta}{d\_3} = 0$$
to
$$\frac{\beta}{d\_3} = \frac{1}{d\_2-d\_3}$$
Flip it:
$$\frac{d\_3}{\beta} = d\_2-d\_3$$
The equation is now linear in $d\_3$.


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