Determining solution of a differential equation

Question

$ Problem$

The graph of a solution $u(x)$ of the differential equation $ y''-4y'+29y=0 $ intersects the graph of a solution $v(x)$ of the equation $y"+4y'+13y=0$ at the origin . Determine $u(x)$ and $v(x)$ if the two curves have equal slopes at the origin and if $u'(\pi/2)= 1$.

$Attempt$

$$u(x) =e^{-2x}( a \cos(5x) + b \sin(5x))$$ $$v(x) =e^{-2x}( c \cos(3x) + d \sin(3x))$$

Using three conditions above( using it in order ) we get

1) $a= c$

2) $5 b + 2 a= 3 d- 2 c$

3)$ 2 b - 5 a = e^{- \pi}$

$ Doubt $

How to proceed further ? There are four unknowns and three equations .

Answer 1

$u,v$ both pass through the origin. So

$$a=0$$

$$c=0$$

separately. There are 4 equations

Answer 2

Solutions intersect at the origin,hence

$u(0) =v(0) = 0. $

Answer 3

From the "origin" condition, you immediately draw

$$u(x)=ae^{-2x}\sin5x,\\v(x)=be^{-2x}\sin3x.$$

Then the given slope at $\dfrac\pi2$ means $-2ae^{-\pi}=1$, and equal slopes at the origin implies $-\dfrac52e^\pi=3b$.