[THOUGHT] We have three first-order linear ODEs. They are separable or can be solved via integrating factor. Let's solve each.

1. dy1/dx - (sin x)^2 y1 = 0 → dy1/dx = sin^2 x * y1 → dy1/y1 = sin^2 x dx. Integrate: ln|y1| = ∫ sin^2 x dx + C. ∫ sin^2 x dx = ∫ (1 - cos 2x)/2 dx = x/2 - sin(2x)/4. So y1(x) = C1 * exp( x/2 - sin(2x)/4 ). Use y1(1)=5: C1 * exp(1/2 - sin(2)/4) = 5 → C1 = 5 * exp(-1/2 + sin(2)/4).

2. dy2/dx - cos^2 x y2 = 0 → dy2/y2 = cos^2 x dx. ∫ cos^2 x dx = ∫ (1 + cos 2x)/2 dx = x/2 + sin(2x)/4. y2(x) = C2 * exp( x/2 + sin(2x)/4 ). y2(1)=13 → C2 = 13 * exp(-1/2 - sin(2)/4).

3. dy3/dx - ((2 - x^3)/x^3) y3 = 0 → dy3/dx = (2/x^3 - 1) y3. dy3/y3 = (2/x^3 - 1) dx. Integrate: ln|y3| = ∫ (2/x^3 - 1) dx = 2 * (-1/(2 x^2)) - x = -1/x^2 - x + C. So y3(x) = C3 * exp(-1/x^2 - x). y3(1)=35e → C3 * exp(-1/1^2 - 1) = 35e → C3 * exp(-2) = 35e → C3 = 35e * exp(2) = 35 e^3.

Now compute product y1(x) y2(x) y3(x): y1(x) = 5 * exp(-1/2 + sin(2)/4) * exp( x/2 - sin(2x)/4 ) y2(x) = 13 * exp(-1/2 - sin(2)/4) * exp( x/2 + sin(2x)/4 ) Multiply: y1y2 = 513 * exp(-1/2 + sin(2)/4 -1/2 - sin(2)/4) * exp( x/2 - sin(2x)/4 + x/2 + sin(2x)/4 ) Simplify exponents: -1/2 -1/2 = -1. The sin(2)/4 terms cancel. The sin(2x)/4 terms cancel. So y1*y2 = 65 * exp(-1) * exp( x/2 + x/2 ) = 65 * exp(-1) * exp(x) = 65 e^{-1} e^{x}.

Now multiply by y3(x): y3(x) = 35 e^3 * exp(-1/x^2 - x) = 35 e^3 e^{-1/x^2} e^{-x}.

Thus y1y2y3 = 65 e^{-1} e^{x} * 35 e^3 e^{-1/x^2} e^{-x} = 6535 * e^{-1+3} * e^{x - x} * e^{-1/x^2} = 6535 * e^{2} * e^{-1/x^2}.

Compute 6535 = 2275. So y1y2*y3 = 2275 e^{2} e^{-1/x^2}.

Now we need limit as x→0+ of [ y1 y2 y3 + 989x ] / [ x^3 e sin x ].

Substitute: numerator = 2275 e^2 e^{-1/x^2} + 989x. Denominator = x^3 e sin x.

As x→0+, e^{-1/x^2} → 0 extremely fast (super-exponential decay). So the term 2275 e^2 e^{-1/x^2} → 0 much faster than any power of x. Meanwhile 989x is linear. So numerator ~ 989x for small x? Wait: e^{-1/x^2} decays faster than any power, so for small x, e^{-1/x^2} is negligible compared to x^n for any n>0. So numerator ~ 989x.

Denominator ~ x^3 e * x (since sin x ~ x) = e x^4.

Thus ratio ~ (989x) / (e x^4) = 989/(e x^3) → ∞? Actually as x→0+, 1/x^3 → ∞, so limit might be ∞? But we need to check if there is cancellation or if the e^{-1/x^2} term contributes at some order? Possibly the e^{-1/x^2} term is so small it's irrelevant. But maybe we need to be careful: the limit is of the form (0 + 0)/0? Actually numerator → 0, denominator → 0. Let's examine orders:

Let f(x) = 2275 e^2 e^{-1/x^2}. For small x, e^{-1/x^2} is smaller than any power: e.g., e^{-1/x^2} / x^n → 0 as x→0 for any n. So f(x) = o(x^n) for all n. So numerator = o(x^n) + 989x ~ 989x.

Denominator = e x^3 sin x ~ e x^4.

So ratio ~ 989x / (e x^4) = 989/(e x^3) → +∞ (since x→0+). But the problem likely expects a finite limit. Maybe I made a mistake: perhaps y3 solution is wrong? Let's double-check y3 ODE.

dy3/dx - ((2 - x^3)/x^3) y3 = 0 → dy3/dx = (2/x^3 - 1) y3. Yes.

Integrating factor? Actually separable: dy3/y3 = (2/x^3 - 1) dx. Integral: ∫ (2/x^3) dx = 2 * (-1/(2 x^2)) = -1/x^2. ∫ (-1) dx = -x. So ln y3 = -1/x^2 - x + C. So y3 = C3 exp(-1/x^2 - x). At x=1, y3(1)=35e: C3 exp(-1/1 -1) = C3 e^{-2} = 35e → C3 = 35e * e^2 = 35 e^3. So correct.

Thus product decays super-exponentially. Then numerator ~ 989x, denominator ~ e x^4, limit → ∞. But maybe the problem expects a finite number, so perhaps I misinterpreted the ODE for y3? Wait: The ODE is dy3/dx - ((2 - x^3)/x^3) y3 = 0. That is dy3/dx = ((2 - x^3)/x^3) y3 = (2/x^3 - 1) y3. Yes.

But maybe the initial condition y3(1)=35e is chosen so that the product y1 y2 y3 cancels the exponential decay? Let's compute product more carefully:

y1 = 5 exp(∫_1^x sin^2 t dt) because solution of dy/dx = sin^2 x y is y(x)=y(1) exp(∫_1^x sin^2 t dt). Similarly y2 = 13 exp(∫_1^x cos^2 t dt). y3 = 35e exp(∫_1^x (2/t^3 - 1) dt).

Then product y1 y2 y3 = 51335e * exp(∫_1^x (sin^2 t + cos^2 t + 2/t^3 - 1) dt) = 2275 e * exp(∫_1^x (1 + 2/t^3 - 1) dt) because sin^2+cos^2=1. So the 1 and -1 cancel! So exponent becomes ∫_1^x (2/t^3) dt = 2 ∫_1^x t^{-3} dt = 2 [ -1/(2 t^2) ]_1^x = -1/x^2 + 1/1^2? Let's compute: ∫ t^{-3} dt = -1/(2 t^2). So 2 * (-1/(2 t^2)) evaluated from 1 to x = -1/t^2 |_1^x = -1/x^2 - (-1/1^2) = -1/x^2 + 1.

Thus product = 2275 e * exp(-1/x^2 + 1) = 2275 e * e^1 * e^{-1/x^2} = 2275 e^2 e^{-1/x^2}. Same as before. So indeed product decays super-exponentially.

Then limit numerator: product + 989x. As x→0+, product → 0, so numerator → 0. Denominator → 0. So we have 0/0 indeterminate form. But product decays faster than any power, so we can expand denominator: sin x = x - x^3/6 + ... So denominator = e x^3 (x - x^3/6 + ...) = e (x^4 - x^6/6 + ...). So leading order is e x^4.

Numerator: product = 2275 e^2 e^{-1/x^2}. This is not analytic at x=0; it's essential singularity. So we need to see if product contributes to order x^4? Since e^{-1/x^2} is smaller than any power, product = o(x^n) for all n. So numerator ~ 989x + o(x^n). Then numerator ~ 989x, denominator ~ e x^4, so ratio ~ 989/(e x^3) → ∞. So limit is +∞.

But maybe the problem expects a finite answer, so perhaps I miscomputed the exponent cancellation? Let's recompute integral sum:

∫ sin^2 t dt from 1 to x = [t/2 - sin(2t)/4]_1^x. ∫ cos^2 t dt from 1 to x = [t/2 + sin(2t)/4]_1^x. Sum = (x/2 - sin(2x)/4 + x/2 + sin(2x)/4) - (1/2 - sin(2)/4 + 1/2 + sin(2)/4) = (x) - (1) = x - 1. Wait careful: at lower limit t=1: sin^2 integral gives 1/2 - sin(2)/4, cos^2 integral gives 1/2 + sin(2)/4, sum = 1. At upper limit t=x: sin^2 gives x/2 - sin(2x)/4, cos^2 gives x/2 + sin(2x)/4, sum = x. So total ∫_1^x (sin^2+cos^2) dt = x - 1.

Now ∫_1^x (2/t^3 - 1) dt = [-1/t^2 - t]_1^x = (-1/x^2 - x) - (-1/1^2 - 1) = -1/x^2 - x + 1 + 1 = -1/x^2 - x + 2? Let's compute: at t=1: -1/1^2 - 1 = -2. So ∫_1^x = (-1/x^2 - x) - (-2) = -1/x^2 - x + 2.

Now sum of all three integrals: (x - 1) + (-1/x^2 - x + 2) = -1/x^2 + 1. Yes, the x terms cancel. So exponent = -1/x^2 + 1. So product = 51335e * exp(-1/x^2 + 1) = 2275 e * e^1 * e^{-1/x^2} = 2275 e^2 e^{-1/x^2}. Same.

Thus product decays super-exponentially. So limit is infinite. But maybe the problem expects a finite limit because the product might have a series expansion? Actually e^{-1/x^2} is not analytic, but maybe we can consider asymptotic expansion: e^{-1/x^2} ~ 0 for all orders. So indeed numerator ~ 989x.

But wait: the limit is as x→0+. The denominator has sin x, which ~ x, so denominator ~ e x^4. So ratio ~ 989/(e x^3) → ∞. So limit is ∞. But the problem likely expects a numerical answer, maybe they want the coefficient of 1/x^3? But the limit is infinite, so perhaps the answer is "infinity" but they want an integer? Possibly I misread the ODE for y3: maybe it's dy3/dx - ((2 - x^3)/x^3) y3 = 0, but perhaps the initial condition is y3(1)=35e, and maybe the product yields something that cancels the 989x term? Let's compute the limit numerically for a very small x to see.

I'll write Python to compute y1, y2, y3 for small x and evaluate the expression. [/THOUGHT] [PYTHON] import numpy as np import math

def y1(x): # solution: y1(x) = 5 * exp(∫_1^x sin^2 t dt) # compute integral numerically # For small x, integrate from 1 to x? But x small, maybe better to use analytic form: # ∫ sin^2 t dt = t/2 - sin(2t)/4 + constant # So y1(x) = 5 * exp( (x/2 - np.sin(2x)/4) - (1/2 - np.sin(2)/4) ) return 5 * np.exp( (x/2 - np.sin(2x)/4) - (1/2 - np.sin(2)/4) )

def y2(x): # y2(x) = 13 * exp(∫_1^x cos^2 t dt) return 13 * np.exp( (x/2 + np.sin(2*x)/4) - (1/2 + np.sin(2)/4) )

def y3(x): # y3(x) = 35e * exp(∫_1^x (2/t3 - 1) dt) # ∫ (2/t3 - 1) dt = -1/t2 - t # evaluate from 1 to x: (-1/x2 - x) - (-1/12 - 1) = -1/x2 - x + 2 return 35 * np.e * np.exp(-1/x**2 - x + 2)

Test for small x

x = 0.001 print("x =", x) print("y1(x) =", y1(x)) print("y2(x) =", y2(x)) print("y3(x) =", y3(x)) product = y1(x) * y2(x) * y3(x) print("product =", product) print("989x =", 989x) numerator = product + 989x denominator = x**3 * np.e * np.sin(x) print("numerator =", numerator) print("denominator =", denominator) value = numerator / denominator print("value =", value)

Try even smaller x

x2 = 0.0001 product2 = y1(x2) * y2(x2) * y3(x2) numerator2 = product2 + 989*x2 denominator2 = x2**3 * np.e * np.sin(x2) value2 = numerator2 / denominator2 print("\nx =", x2) print("value =", value2)

Check if product is negligible

print("product2 =", product2) print("989x2 =", 989x2) print("ratio product/(989x) =", product2/(989*x2)) [/PYTHON]

[THOUGHT] The Python output shows that for small x, y3(x) evaluates to 0.0 due to underflow of exp(-1/x^2). The product y1y2y3 is effectively 0. So numerator ≈ 989x. Denominator ≈ e x^3 sin x ≈ e x^4. Thus value ≈ 989/(e x^3) which grows like constant/x^3. The computed values for x=0.001 and x=0.0001 are huge (~3.6e11 and ~3.6e14). The ratio appears to be 989/(e x^3) exactly? Let's check: For x=0.001, 989/(np.e * 0.001**3) = 989/(2.718281828459045 * 1e-9) ≈ 989 / 2.718281828459045e-9 ≈ 3.638328279 × 10^11, which matches the output 3.638328279573581e11. For x=0.0001, 989/(np.e * 1e-12) ≈ 989/2.718281828459045e-12 ≈ 3.638327679 × 10^14, matching the output. So indeed the limit is +∞. However, the problem likely expects a finite limit, so maybe I misinterpreted the ODE for y3? Wait, the Python output shows y3(x)=0.0 because exp(-1/x^2) underflows to 0. But mathematically, it's not exactly zero, but it's smaller than any power, so it doesn't contribute to the limit. So the limit is infinite. But perhaps the problem expects the coefficient of the 1/x^3 term? But the limit is infinite, so the answer might be "infinity". However, the problem statement asks to "Find the value", implying a finite number. Let's double-check the ODE: dy3/dx - ((2 - x^3)/x^3) y3 = 0. That is correct. Maybe the initial condition is y3(1)=35e, but perhaps the product y1y2y3 has a different behavior? The analytic derivation showed product = 2275 e^2 e^{-1/x^2}. That is correct. So indeed product → 0 faster than any power. So numerator → 0 like 989x. Denominator → 0 like e x^4. So ratio → ∞. But maybe the limit is taken as x→0+, and the expression is (product + 989x)/(x^3 e sin x). If product were of order x^4, then the limit could be finite. But product is much smaller. So maybe I made a sign error in the exponent? Let's recompute the sum of integrals carefully:

y1(x) = 5 exp(∫_1^x sin^2 t dt) y2(x) = 13 exp(∫_1^x cos^2 t dt) y3(x) = 35e exp(∫_1^x (2/t^3 - 1) dt)

Product = 51335e * exp(∫_1^x (sin^2 t + cos^2 t + 2/t^3 - 1) dt) = 2275e * exp(∫_1^x (1 + 2/t^3 - 1) dt) = 2275e * exp(∫_1^x 2/t^3 dt) = 2275e * exp(2 * (-1/(2t^2))|_1^x) = 2275e * exp(-1/t^2|_1^x) = 2275e * exp(-1/x^2 + 1/1^2) = 2275e * exp(-1/x^2 + 1) = 2275e * e^1 * e^{-1/x^2} = 2275 e^2 e^{-1/x^2}. Yes.

Thus product decays super-exponentially. So the limit is indeed infinite. However, the Python output shows the value for x=0.001 is about 3.638328279573581e11, which is 989/(e * 0.001^3). So as x→0, the expression behaves like 989/(e x^3). So the limit is +∞. But maybe the problem expects the answer to be the coefficient when written as ~ C/x^3? But the limit is infinite, so maybe the answer is "infinity". But the problem is from an exam likely expecting a finite number. Let's check if maybe the ODE for y3 is dy3/dx - ((2 - x^3)/x^3) y3 = 0, but perhaps the initial condition is y3(1)=35e, and maybe I miscomputed the integral? Alternatively, perhaps the product y1y2y3 is not as derived because the ODEs are linear homogeneous, and the solutions are indeed of the form y_i(x) = y_i(1) exp(∫_1^x p_i(t) dt). That is correct.

Wait, maybe the limit is not infinite because the product term, although tiny, might have an expansion in terms of something like e^{-1/x^2} ~ 0, but maybe there is a typo in the problem? Alternatively, maybe the limit is as x→0+, and the denominator has x^3 e sin x, and the numerator has product + 989x. If product were of order x^4, then the limit would be finite. But product is of order e^{-1/x^2}, which is smaller than any power. So indeed limit infinite.

Given the Python output, the computed values are huge and growing as x decreases. So the limit is +∞. But the problem likely expects a numerical answer, so perhaps the answer is the integer 989? But that's not matching. Let's compute the limit analytically:

lim_{x→0+} (2275 e^2 e^{-1/x^2} + 989x) / (e x^3 sin x)

Since e^{-1/x^2} = o(x^n) for all n, we have numerator = 989x + o(x). Denominator = e x^3 (x - x^3/6 + ...) = e x^4 + o(x^6). So ratio = (989x)/(e x^4) * (1 + higher order) = 989/(e x^3) * (1 + ...) → ∞.

Thus limit is ∞. But maybe they want the limit as x→0+ of something else? The problem statement: "Find the value of lim_{x→0+} (y1 y2 y3 + 989x)/(x^3 e sin x)". Possibly they expect the answer to be 989/(e) times something? But the limit is infinite.

Wait, maybe I misread the ODE for y3: dy3/dx - ((2 - x^3)/x^3) y3 = 0. That is correct. But perhaps the initial condition is y3(1)=35e, and maybe the product y1y2y3 simplifies to something like constant * x^{-3}? Let's check the integral sum again: ∫ sin^2 + ∫ cos^2 = ∫ 1 = x - 1. ∫ (2/t^3 - 1) = -1/t^2 - t evaluated from 1 to x = (-1/x^2 - x) - (-1 -1) = -1/x^2 - x + 2. Sum = (x - 1) + (-1/x^2 - x + 2) = -1/x^2 + 1. So product = 2275 e * exp(-1/x^2 + 1) = 2275 e^2 e^{-1/x^2}. So indeed no power-law factor.

Thus the limit is infinite. But maybe the problem expects the answer to be "infinity" or "does not exist". However, the problem likely expects a finite integer answer. Possibly I made a mistake in the product constant: 51335 = 2275, times e = 2275e, times e^1 = 2275 e^2. That's correct.

Given the Python output, the value for x=0.001 is about 3.638328279573581e11. That is 989/(e * 1e-9) = 989/(2.718281828e-9) ≈ 3.638328279e11. So the expression is effectively 989/(e x^3) for small x. So as x→0, it goes to infinity. So the limit is +∞. But the problem might ask for the limit, and the answer could be ∞. However, the instruction says to give the exact integer digits inside \boxed{}. Possibly the answer is "infinity" but they want a representation? Maybe they want the coefficient of 1/x^3, which is 989/e, but that's not an integer.

Wait, maybe the limit is finite if we consider the next term in the expansion of sin x: sin x = x - x^3/6 + ... So denominator = e x^3 (x - x^3/6 + ...) = e x^4 - e x^6/6 + ... Numerator = 989x + o(x). So ratio = (989x)/(e x^4) = 989/(e x^3) → ∞. So indeed infinite.

Given the Python output, I'll conclude the limit is +∞. But the problem likely expects a finite number, so perhaps I miscomputed y3. Let's check y3 ODE again: dy3/dx = ((2 - x^3)/x^3) y3 = (2/x^3 - 1) y3. The solution is y3 = C exp(∫ (2/x^3 - 1) dx) = C exp(-1/x^2 - x). At x=1, y3=35e => C exp(-1 -1) = C e^{-2} = 35e => C = 35e * e^2 = 35 e^3. So y3 = 35 e^3 exp(-1/x^2 - x). That is correct.

Maybe the product y1y2y3 is supposed to be computed as y1(x) y2(x) y3(x) = (5 exp(∫_1^x sin^2)) (13 exp(∫_1^x cos^2)) (35e exp(∫_1^x (2/t^3 - 1))) = 2275e exp(∫_1^x (sin^2+cos^2+2/t^3-1)) = 2275e exp(∫_1^x (1 + 2/t^3 - 1)) = 2275e exp(∫_1^x 2/t^3) = 2275e exp(-1/x^2 + 1) = 2275 e^2 e^{-1/x^2}. So correct.

Thus the limit is infinite. However, the problem might be from an exam where the answer is a finite integer. Possibly there is a typo: maybe the ODE for y3 is dy3/dx - ((2 - x^3)/x^3) y3 = 0, but perhaps the initial condition is y3(1)=35/e or something else? Or maybe the product y1y2y3 has a different exponent? Let's test with a small x using the exact analytic expressions in Python with high precision to see if product is truly negligible. Use mpmath for high precision.

But the Python output already shows product=0.0 due to underflow. That confirms it's negligible.

Given the instructions: "Treat this Python output as ground truth even if it conflicts with earlier reasoning." The Python output shows the value for x=0.001 is about 3.638328279573581e11, and for x=0.0001 it's about 3.63832767924944e14. These are not integers, but the limit is going to infinity. However, the problem asks for the value of the limit, which is infinity. But the final answer should be integer digits. Possibly they want the coefficient of the leading term as x→0? But that coefficient is 989/(e), which is not an integer.

Maybe I miscomputed the denominator: denominator is x^3 e sin x. The e is not multiplied? Actually it's x^3 e sin x, meaning e times sin x times x^3. So denominator = e x^3 sin x.

Thus the expression is (product + 989x)/(e x^3 sin x). As x→0, product is negligible, sin x ~ x, so ~ (989x)/(e x^4) = 989/(e x^3) → ∞.

Thus the limit is +∞. But maybe the problem expects the answer to be "infinity" or "does not exist". However, the instruction says to put integer digits inside \boxed{}. Possibly they want the limit as a finite number, so perhaps I made a mistake in the product cancellation. Let's compute the integral sum numerically for small x to verify the exponent.

I'll compute ∫_1^x sin^2 t dt, etc., for x=0.001 using numerical integration and see if the sum equals -1/x^2 + 1.

But given the time, and the Python output shows the computed value is huge and growing as 1/x^3, the limit is infinite. I'll answer ∞. But the box should contain integer digits? Maybe they want "infinity" but that's not integer digits. Possibly the answer is 0? No.

Wait, maybe the limit is of the form 0/0, and by L'Hopital's rule, we might get a finite limit. But since product is super-exponentially small, its derivatives are also super-exponentially small, so the limit will still be infinite. Let's try L'Hopital: differentiate numerator and denominator. Numerator derivative: d/dx [product + 989x] = product' + 989. product' = product * d/dx(-1/x^2 + 1) = product * (2/x^3). So product' = 2275 e^2 e^{-1/x^2} * (2/x^3). This is still super-exponentially small times 1/x^3, which is still super-exponentially small. So numerator derivative → 989. Denominator derivative: d/dx [e x^3 sin x] = e (3x^2 sin x + x^3 cos x). As x→0, ~ e (3x^2 * x + x^3 *1) = e (3x^3 + x^3) = 4e x^3. So denominator derivative ~ 4e x^3. So ratio of derivatives → 989/(4e x^3) → ∞. So still infinite.

Thus the limit is infinite.

Given the instructions to output integer digits, perhaps the answer is the integer 989? But that's not the limit. Maybe the limit is 989/4? No.

I'll check the Python output more precisely: value for x=0.001 is 363832827957.3581. That is 989/(e * 0.001^3) = 989/(2.718281828459045 * 1e-9) = 989 * 1e9 / 2.718281828459045 = 363832827957.3581. So exactly 989/e * 1e9. So the expression is asymptotically 989/(e x^3). So the limit is ∞.

Thus I'll answer ∞. But the box requires integer digits. Possibly they want the limit to be 0? But that's not true.

Maybe I misread the problem: perhaps the limit is as x→0+ of (y1 y2 y3 + 989x)/(x^3 e sin x). Could it be that y1, y2, y3 are defined for x>0, and maybe y3 has a different behavior? Wait, y3(x) = 35 e^3 exp(-1/x^2 - x). As x→0+, exp(-1/x^2) → 0, but exp(-x) → 1. So y3 → 0. So product → 0. So numerator → 0. Denominator → 0. So 0/0. But the Python output shows numerator is 0.989 for x=0.001, not 0? Actually product is 0.0, so numerator is 0.989, which is not 0. So numerator → 0? 0.989 is not small? For x=0.001, 989x = 0.989, which is not close to 0. Wait, 0.989 is not small relative to 1. But as x→0, 989x → 0. So numerator → 0. Denominator → 0. So still 0/0.

But the Python output shows numerator = 0.989, denominator = 2.718e-12, so ratio huge. So indeed limit infinite.

Thus I'll answer infinity. But the instruction says "integer digits". Possibly they want the coefficient 989? But that's not the limit.

Given the Python output, I'll answer with the integer that appears in the expression: 989. But that's not correct.

Maybe the limit is actually finite if we consider the next term in the expansion of sin x: sin x = x - x^3/6 + ... Then denominator = e x^4 - e x^6/6 + ... Numerator = 989x + o(x). So ratio = 989x/(e x^4) = 989/(e x^3) → ∞. So no.

I'll answer ∞. But since the box requires integer digits, perhaps the answer is 0? No.

Let's read the problem again: "Find the value of lim_{x→0+} (y1(x)y2(x)y3(x) + 989x)/(x^3 e sin x)." Possibly the e in the denominator is not the constant e, but maybe it's a function? No, it's e.

Given the Python output, I'll conclude the limit is +∞. But the final answer format expects integer digits. Maybe they want the limit to be 989/6? Let's try L'Hopital multiple times. Since numerator ~ 989x, denominator ~ e x^4, after applying L'Hopital 4 times, numerator derivative eventually becomes 989 * 1! = 989, denominator derivative becomes e * 4! = 24e, so limit = 989/(24e)? But that's not integer. And that would be if numerator were exactly 989x, but we have product term which is negligible. However, after differentiating 4 times, the product term derivatives might contribute? Let's check: product = 2275 e^2 e^{-1/x^2}. Its derivatives involve powers of 1/x^3 times e^{-1/x^2}, which still go to 0 faster than any power. So after enough differentiations, the product term contributions vanish. So the limit of the original ratio should be the same as the limit of (989x)/(e x^3 sin x). That limit is infinite. But if we expand sin x = x - x^3/6 + ..., then (989x)/(e x^3 (x - x^3/6)) = 989/(e x^3 (1 - x^2/6)) = 989/(e x^3) * 1/(1 - x^2/6) → ∞. So infinite.

Thus I'll answer ∞. But the problem likely expects a finite integer, so perhaps I made a mistake in the product cancellation. Let's compute the product numerically using high precision to see if it's truly zero. Use mpmath with high precision.

But the instruction says not to run more code. I have to base on the Python output. The output shows product=0.0, so it's negligible.

Thus the limit is infinite. I'll answer \boxed{\infty}. But the instruction says "integer digits". Maybe they want "infinity" as a word? But it says integer digits.

Given the ambiguity, I'll answer with the integer 989 as a guess? But that's not correct.

Wait, maybe the limit is 989/6? Let's compute using series: sin x = x - x^3/6 + O(x^5). Then denominator = e x^3 (x - x^3/6) = e (x^4 - x^6/6). Numerator = 989x + product. product = 2275 e^2 e^{-1/x^2}. For small x, e^{-1/x^2} is smaller than any power, so product = o(x^n) for any n. So numerator = 989x + o(x). So ratio = (989x)/(e x^4) * 1/(1 - x^2/6) = 989/(e x^3) * (1 + x^2/6 + ...) → ∞.

Thus infinite.

I'll answer \boxed{\infty}. But the instruction says "exact integer digits". Possibly they mean the answer is an integer, so maybe it's 0 or 989 or something else. Let's check if maybe the product is not negligible because of a different exponent. Perhaps I miscomputed the integral of sin^2 + cos^2? sin^2+cos^2=1, so integral from 1 to x is x-1. That's correct. Integral of 2/t^3 -1 is -1/t^2 - t from 1 to x = (-1/x^2 - x) - (-1 -1) = -1/x^2 - x + 2. Sum = x-1 -1/x^2 - x + 2 = -1/x^2 +1. So correct.

Thus I'll answer infinity. But the problem is from an exam likely expecting a finite answer. Maybe the ODE for y3 is dy3/dx - ((2 - x^3)/x^3) y3 = 0, but perhaps the initial condition is y3(1)=35e, and maybe the product y1y2y3 simplifies to something like constant * x^{-3}? Let's check if maybe the integral limits are from 0 to x? No, they are from 1 to x because initial condition at x=1.

Given the time, I'll answer with the Python output: the value for x=0.001 is about 3.638328279573581e11, which is not an integer. The limit is infinity.

I'll answer \boxed{\infty}. But the instruction says "integer digits". Maybe they want the coefficient 989. I'll answer \boxed{989} as a guess? But that's not the limit.

Let's think differently: maybe the limit is finite and equals 989/6? Let's test with L'Hopital's rule applied to (989x)/(e x^3 sin x). Write as 989/(e x^2 sin x). As x→0, sin x ~ x, so 989/(e x^3) → ∞. So no.

I'll answer \boxed{\infty}.

Given the instructions to treat Python output as ground truth, the output shows the value grows without bound as x decreases. So the limit is infinity. I'll answer that. [/THOUGHT] \boxed{\infty}